Loci | ICSE Class 10 Maths Notes

This note covers the meaning of loci, distance conditions, the three basic locus theorems, ruler-and-compass constructions, and finding the centres of circles through given points.

What is a locus, and what does its condition tell us?

Definition: A locus is the set of all points that satisfy a stated condition. Loci is the plural of locus. A set means a collection of objects, here points.

A point represents an exact position. A plane is a flat surface extending in all directions. The loci considered here lie in a plane, like the geometrical figures represented on a flat sheet of paper.

A fixed point has an unchanged position. A locus question gives a condition and asks which positions satisfy it. The condition might require a fixed distance from a given point, equal distances from two given points, or equal distances from two intersecting lines.

Why must the answer include all suitable points?

The word all matters. Finding a single position that satisfies the condition does not necessarily identify the whole locus. A complete answer describes the entire collection of suitable positions, while excluding positions that fail the condition.

There are two checks. Every point on the proposed locus must satisfy the condition. Conversely, every point satisfying the condition must belong to the proposed locus. Conversely means considering the implication in the reverse direction.

This is why a few correctly plotted points are useful for recognising a shape but do not, by themselves, establish its complete extent. The relevant locus theorem identifies the full shape and connects it to the given condition.

A theorem is a mathematical statement established by reasoning. Here, the three basic theorems let us translate a distance condition into its corresponding geometrical locus. Their wording supplies the reason for a construction.

The expression moving point describes a point whose permitted position can vary. No speed or time is involved. We are interested in where the point may be, rather than how quickly it travels between positions.

How are the distances in a locus condition measured?

The distance between two points is the length of the line segment joining them. A line segment has two endpoints. A straight line extends indefinitely in both directions, while a ray has one endpoint and extends indefinitely in one direction.

Letters name points. If A and B are points, AB can name the segment joining them or its length, according to context. In an equality between lengths, AB means the length. The symbol = means “is equal to”.

What does equidistant mean?

Equidistant means at equal distances. If P is a point whose distances from A and B are equal, we write PA = PB. This compares two lengths measured from the same point P.

The common distance need not be specified. Different points satisfying PA = PB can have different common distances from A and B. By contrast, a fixed distance has one given value which stays unchanged throughout the condition.

How is distance from a line different?

An angle measures the opening between two rays with a common endpoint, called the vertex. A right angle measures 90°, where ° denotes degrees, a unit of angle. Lines meeting at a right angle are perpendicular.

The distance from a point to a line is measured along the perpendicular from the point to that line. The point where this perpendicular meets the line is its foot. A slanting segment to an arbitrary point on the line does not measure this distance.

Thus “equidistant from two lines” compares two perpendicular lengths. It does not compare distances to the intersection of the lines or to selected points marked on them. Two intersecting lines are lines sharing a common point.

Note: First identify what the distances are measured from. Two given points lead to a different locus from two intersecting lines, even though both conditions use the word equidistant.

What is the locus at a fixed distance from a fixed point?

Theorem: Fixed distance from a fixed point

The locus of a point in a plane at a fixed positive distance from a fixed point is a circle. The fixed point is its centre, and the fixed distance is its radius, the distance from the centre to any point on the circle.

“Positive” means greater than zero. This qualification makes explicit that we are describing a circle with a non-zero radius. The centre is fixed, and the radius stays unchanged as we consider different points on the circle.

The circle consists of its boundary points. Its interior, meaning the region inside that boundary, is not included in this equality condition. Points inside are nearer the centre than the radius; points outside are farther from it.

How do we specify and draw this circle?

A complete description names both centre and radius. Merely writing “a circle” leaves its position and size undetermined. A compass is the drawing instrument whose fixed opening allows a pencil to trace points at a constant distance from its anchored point.

Place the compass point at the given centre, set its opening to the given distance, and draw the full circle. Keep the opening unchanged throughout the turn. Every traced point then has the required distance from the fixed centre.

Worked example 1. A is a given point and B is a different given point. Describe the locus of points whose distance from A equals the length AB.

Answer: The locus is the circle with centre A and radius AB. Set the compass opening to AB and draw the circle about A. B lies on it because its distance from A is the specified radius.

The circle has centre A, with B and C on its boundary. Segments AB and AC join the centre to the boundary; segment BC joins two boundary points. D and E are also labelled on the circle.

See Fig. 5.3 in your NCERT textbook

What is the locus equidistant from two given points?

A midpoint divides a segment into two equal parts. To bisect means to divide into two equal parts. The perpendicular bisector of a segment is the straight line through its midpoint, perpendicular to that segment.

Theorem: Equal distances from two points

The locus of a point equidistant from two given distinct points is the perpendicular bisector of the segment joining them. Distinct points occupy different positions. Their separation determines the segment whose perpendicular bisector is required.

Let A and B be the given points, and let M be the midpoint of AB. Thus AM = MB. The required line passes through M and meets AB at a right angle. The entire line is the locus, not just M.

Every point on this line is equidistant from A and B. Conversely, every point equidistant from A and B lies on this line. Both directions matter: together they show that the perpendicular bisector contains all and only the required points.

Why is a line through the midpoint not enough?

A line can pass through the midpoint without being perpendicular to the segment. Such a line does not in general give the required locus. Equally, a perpendicular drawn through another point of the segment does not bisect it.

Use both parts of the name as a check: perpendicular checks the angle, and bisector checks the division of the segment. The given points themselves do not lie on their perpendicular bisector because they are distinct.

Worked example 2. A and B are distinct fixed points in a plane. Identify the locus of a point P for which PA = PB, and explain whether the midpoint alone is sufficient.

Answer: Construct the perpendicular bisector of AB. It is the complete locus. The midpoint satisfies the condition, but so does every other point on this perpendicular bisector; therefore the midpoint alone is incomplete.

The common distance from P to A and B may change as P takes different positions on the line. Equal distances should therefore not be confused with one prescribed fixed distance.

What is the locus equidistant from two intersecting lines?

An angle bisector divides an angle into two equal angles. When two straight lines intersect, they form angles around their common point. The bisectors of these angles form the locus associated with equal perpendicular distances from the lines.

Theorem: Equal distances from intersecting lines

The locus of a point equidistant from two intersecting lines is the bisectors of the angles between the lines. For the complete lines, the full locus consists of both angle-bisector lines, rather than a single ray inside one chosen angle.

If the point is required to lie within one specified angle, retain the bisector ray in that angle. The words describing the permitted region are part of the condition. A locus must satisfy the region restriction as well as the equality of distances.

How do we check the condition?

Choose a point P on the relevant bisector. Let H and K be the feet of the perpendiculars from P to the two lines. The required equality is PH = PK, where PH and PK are the perpendicular lengths.

The distances being compared are not lengths measured along the two lines from their intersection. Nor is the condition a fixed distance from the vertex. Those interpretations replace the stated condition with a different one.

Worked example 3. Two given straight lines intersect. A point may lie anywhere in their plane and must be equidistant from the lines. Identify the complete locus and say how its distances are measured.

Answer: Draw both angle-bisector lines through the intersection. Together they form the complete locus. The distance to each given line is measured along a perpendicular from the moving point to that line, and these perpendicular lengths are equal.

Draw two intersecting straight lines and their angle bisectors. Mark a point on one bisector and drop perpendiculars to the given lines. Mark the perpendicular lengths equal and show the right angles at their feet.

The distinction between the full locus and a restricted part prevents a common incomplete answer. Read whether the question refers to whole intersecting lines or confines the point to one particular angle before choosing what to retain.

How are the basic loci constructed accurately?

A construction is a geometrical drawing made by a specified sequence of steps. A ruler provides a straight edge for drawing lines; a compass transfers lengths and draws circles or arcs. An arc is a part of a circle's boundary.

How do we construct a perpendicular bisector?

For distinct given points A and B, start with the segment AB. The construction uses equal compass openings from its endpoints to obtain points equidistant from them. These points determine the required straight line.

1. Set the compass opening to more than half the length of AB, so that equal arcs from its endpoints can cross on both sides of the segment.
2. With A as centre, draw arcs on both sides of AB. Keep this compass opening unchanged for the next step.
3. With B as centre, draw arcs cutting the first pair. Name the two arc intersections C and D.
4. Draw the straight line through C and D and extend it in both directions. This is the perpendicular bisector of AB and hence the required locus.

Each arc intersection is equally distant from A and B because both arcs use the same radius. The locus theorem therefore places C and D on the perpendicular bisector. Drawing the line through these distinct intersections locates it.

How do we construct an angle bisector?

Let O be the vertex of the angle to be bisected. Begin with an arc centred at O. The points where it cuts the two arms, meaning the rays forming the angle, will be called A and B.

1. Draw an arc centred at O that cuts the two arms at A and B. The lengths OA and OB are equal.
2. Using equal compass openings, draw arcs centred at A and B that meet inside the angle at a point P different from O.
3. Draw the ray OP. It bisects the chosen angle into two equal angles.
4. For the locus relative to two complete intersecting lines, also bisect an adjacent angle and extend the bisectors to obtain the complete pair of lines.

Adjacent angles here share an arm and the vertex and lie next to one another. The additional construction supplies the other angle-bisector line needed when the point is unrestricted in the plane.

Note: Retain the construction arcs, label their intersections clearly, and distinguish them from the final locus. An arc used to locate a bisector is an auxiliary construction, meaning a helping part of the drawing.

How do we select a locus and combine conditions?

Translate the condition before starting the drawing. Identify the fixed objects, identify which distances must be equal or fixed, and then select the appropriate theorem. This avoids choosing a familiar shape simply because the word “distance” occurs.

What if the same point must satisfy two conditions?

Construct a locus for each condition. The required points belong to both loci, so they must be at their intersection, meaning their common points. Check each common point against the full wording, including any stated restriction on its position.

In a circle-centre construction through three given points, the centre must be equidistant from one pair and also from another pair. Two perpendicular bisectors express these requirements. Their common point supplies the centre, provided the three given points are not on one straight line.

There is a distinction between drawing all possible positions for one condition and identifying positions satisfying all conditions together. The first stage may produce a complete line; the combined conditions may select a particular point on it.

What should a finished answer explain?

1. Name the locus for each stated condition, including its centre and radius or the segment or angle being bisected.
2. Carry out the corresponding construction and keep enough auxiliary marks to show how it was obtained.
3. Identify the common point or points if more than one condition is imposed on the same unknown point.
4. State why the selected result satisfies every condition, rather than giving a label without its geometrical reason.

A labelled diagram and a short explanation work together. The diagram shows where the result lies; the theorem explains why that position has the required property. A measurement can check accuracy, but the theorem supplies the general reason.

How do loci locate circles through given points?

Where can the centre lie for two given points?

A circle through distinct points A and B has a centre O satisfying OA = OB. Its centre must therefore lie on the perpendicular bisector of AB. Conversely, any point on that bisector can serve as the centre of a circle through both points.

Worked example 4. Two distinct points A and B are given in a plane. Where are the centres of all circles passing through both, and how many such circles are possible?

Answer: The centres lie on the perpendicular bisector of AB. Choose any point on that line and use its distance from A as the radius; its equal distance from B places B on the same circle. Infinitely many such circles are possible.

The left group shows circles through A and B. The right group shows circles through C and D, with centres K, J and L marked on the same dashed perpendicular bisector.

See Fig. 5.4 in your NCERT textbook

Worked example 5. A and B are distinct given points. What is the least possible radius of a circle passing through both?

Answer: The least radius is AB/2, where / denotes division. Use the midpoint of AB as centre. AB is then a diameter, a segment joining two points of the circle through its centre. A diameter has twice the radius.

Here the expression AB/2 uses the given separation; it does not require an assumed numerical length. Choosing another point on the perpendicular bisector gives a larger radius because its distance to either endpoint exceeds half the separation.

How does a third point determine the centre?

Three points are collinear if they lie on one straight line, and non-collinear if they do not. Through three distinct non-collinear points there is a unique circle, meaning exactly one circle. There is no circle through three distinct collinear points.

For non-collinear A, B and C, construct the perpendicular bisectors of AB and AC. Call their intersection O. The first gives OA = OB; the second gives OA = OC. Hence O is equally distant from A, B and C.

A triangle is a plane figure formed by three segments joining three non-collinear points. Its vertices are the points where its sides meet. Draw the circle with centre O and radius OA; it passes through all three vertices.

The circumcircle of a triangle is the circle through its vertices; its centre is the circumcentre. Thus constructing the perpendicular bisectors locates the circumcentre and provides the radius needed to draw the circumcircle.

Triangle ABC has its three vertices on the circle. The perpendicular bisectors of its sides are drawn and labelled, meeting at O inside the triangle.

See Fig. 5.5 in your NCERT textbook

How are loci used in complete triangle constructions?

A triangle's interior angles are the angles inside it at its vertices. The angle notation ∠A denotes the angle at vertex A; ∠ means “angle”.

An acute angle is less than 90°, and an obtuse angle is greater than 90° but less than 180°. In an acute-angled triangle all angles are acute; an obtuse-angled triangle has an obtuse angle. A right-angled triangle has a right angle.

The circumcentre lies inside an acute-angled triangle, outside an obtuse-angled triangle, and at the midpoint of the hypotenuse, the side opposite the right angle, in a right-angled triangle. The construction method still uses perpendicular bisectors.

How do we use a side and two angles?

Worked example 6. Construct triangle ABC with AB = 5 cm, ∠A = 70° and ∠B = 60°, and draw its circumcircle. Here cm denotes centimetres, a unit of length. State whether the centre lies inside or outside the triangle.

Answer: Draw AB = 5 cm. At A and B construct rays making the given interior angles on the same side of AB; their meeting point is C. The angle at C is 180° − 70° − 60° = 50°, where − means subtraction. All three angles are acute.

Construct the perpendicular bisectors of AB and AC, meeting at O. Draw the circle with centre O and radius OA. It passes through B and C because OA = OB = OC. The circumcentre lies inside this acute-angled triangle.

The angle calculation uses the fact that a triangle's interior angles add to 180°. A protractor, an instrument for measuring and drawing angles, sets out the given angles; the perpendicular bisectors then locate the circle's centre.

How do we use two sides and their included angle?

Worked example 7. Construct triangle ABC with AB = 5 cm, ∠A = 100° and AC = 4 cm. Draw its circumcircle and state the position of its centre.

Answer: Draw AB = 5 cm. At A draw a ray making 100° with AB, and mark C on it so that AC = 4 cm. Join B to C. Construct perpendicular bisectors of AB and AC, extending them to their intersection O.

Since OA = OB and OA = OC, the circle with centre O and radius OA passes through all three vertices. Its centre lies outside the triangle because the angle at A is obtuse. Leave room outside the triangle for the bisectors to meet.

The included angle is the angle between the two given sides. Here those sides meet at A, so the specified angle fixes their relative directions. An outside centre is a valid result, rather than a reason to redraw it inside.

How do we use three given sides?

Worked example 8. Construct triangle ABC with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw its circumcircle and explain how to check the distances from its centre to the vertices.

Answer: Draw AB = 6 cm. Draw arcs of radius 7 cm centred at A and B, and choose an intersection as C. Join AC and BC. Construct the perpendicular bisectors of AB and AC; call their intersection O.

Draw the circle with centre O and radius OA. Measure OA, OB and OC using the same unit: their lengths should agree within drawing accuracy. The exact geometrical relation is OA = OB = OC, since O lies on the two perpendicular bisectors.

A drawn measurement depends on the precision of the construction. The equality of these radii is the mathematical conclusion; a particular measured decimal radius is not needed to establish that the circle passes through all three vertices.

How can coordinates check a locus?

The distance formula translates a distance condition into an equation. For points $P(x_1,y_1)$ and $Q(x_2,y_2)$, it gives $PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Coordinates use the same length unit on both axes.

For an equal-distance condition, equate the squared distances and simplify. Distances are non-negative, so equality of their squares also ensures equality of the distances themselves.

Worked example 9. Find a relation between $x$ and $y$ such that $P(x,y)$ is equidistant from $A(7,1)$ and $B(3,5)$.

Answer: Set $PA^2=PB^2$. The distance formula gives $(x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2$.

Expanding gives $x^2-14x+49+y^2-2y+1=x^2-6x+9+y^2-10y+25$. Cancel the squared terms and collect the remaining terms: $-8x+8y+16=0$.

Dividing by $-8$ gives $x-y=2$. This line is the perpendicular bisector of $AB$, so it is the complete locus of points equidistant from the two given points.

Worked example 10. Find the point on the $y$-axis equidistant from $A(6,5)$ and $B(-4,3)$.

Answer: A point on the $y$-axis has first coordinate $0$, so write $P(0,y)$. Equating squared distances gives $(6-0)^2+(5-y)^2=(-4-0)^2+(3-y)^2$.

Expand to obtain $36+25-10y+y^2=16+9-6y+y^2$. Cancel $y^2$, giving $61-10y=25-6y$. Hence $36=4y$, so $y=9$ and the required point is $(0,9)$.

Check: $AP=\sqrt{36+16}=\sqrt{52}$ units and $BP=\sqrt{16+36}=\sqrt{52}$ units. Thus both distances agree. The point is the intersection of the $y$-axis and the perpendicular bisector of $AB$.

Worked example 11. Find the values of $y$ for which the distance between $P(2,-3)$ and $Q(10,y)$ is $10$ units.

Answer: Squaring the distance formula gives $(10-2)^2+(y-(-3))^2=10^2$. Thus $64+(y+3)^2=100$, so $(y+3)^2=36$.

Taking both square roots gives $y+3=6$ or $y+3=-6$. Therefore $y=3$ or $y=-9$, giving the points $(10,3)$ and $(10,-9)$.

Check each distance: $\sqrt{(10-2)^2+(3+3)^2}=\sqrt{64+36}=10$ units and $\sqrt{(10-2)^2+(-9+3)^2}=\sqrt{64+36}=10$ units. Both values satisfy the required fixed distance.

Worked example 12. Find a relation between $x$ and $y$ such that $P(x,y)$ is equidistant from $A(3,6)$ and $B(-3,4)$.

Answer: Equate squared distances: $(x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2$.

Expanding gives $x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16$. Cancel the squared terms and collect terms: $-12x-4y+20=0$.

Divide by $-4$ to obtain $3x+y-5=0$, or $3x+y=5$. Reversing these steps gives equal squared distances, hence equal distances. The relation describes the perpendicular bisector of $AB$.

Glossary

• Locus — The set of all points satisfying a stated geometrical condition in the permitted region.
• Plane — A flat surface extending in all directions, containing the points and figures considered here.
• Equidistant — At equal distances from the specified objects, using perpendicular distances when the objects are lines.
• Circle — The set of points in a plane at a fixed positive distance from its centre.
• Radius — The distance from the centre of a circle to any point on its boundary.
• Midpoint — The point dividing a line segment into two parts of equal length.
• Perpendicular bisector — A straight line passing through a segment's midpoint and meeting the segment at a right angle.
• Angle bisector — A ray or line dividing the relevant angle into two equal angles.
• Perpendicular distance — The length of the perpendicular segment from a point to the specified line.
• Intersection — A point belonging to both of the lines or curves being considered.
• Collinear points — Points lying on the same straight line, rather than determining a triangle.
• Circumcircle — The circle passing through all three vertices of a given triangle.
• Circumcentre — The centre of a triangle's circumcircle, found at the intersection of its perpendicular bisectors.

Common errors and misconceptions

• Misconception: One point satisfying a condition is its complete locus. Correct: Include every point satisfying the condition and exclude points that fail it.
• Misconception: A fixed-distance circle includes its interior. Correct: The equality condition selects the boundary; interior points have smaller distances from the centre.
• Misconception: Equal distances from two points mean a fixed radius. Correct: The distances must agree at each position, but their common value can change along the perpendicular bisector.
• Misconception: Any line through a segment's midpoint is its perpendicular bisector. Correct: It must also meet the segment at a right angle.
• Misconception: Distance from a line can be measured along any joining segment. Correct: Use the perpendicular from the point to that line.
• Misconception: One angle-bisector ray gives the unrestricted locus for two intersecting lines. Correct: The full locus contains both angle-bisector lines.
• Misconception: A circumcentre must lie inside its triangle. Correct: It lies outside an obtuse-angled triangle and at the hypotenuse's midpoint in a right-angled triangle.

Exam-style questions with model answers

Q1. Define a locus and explain why one point satisfying a condition need not be a complete answer. [2 marks]

1. A locus is the set of all points that satisfy a stated condition.
2. A single suitable point may omit other positions satisfying the same condition; the complete locus includes them all.

Q2. A and B are distinct given points in a plane. Describe the locus of points whose distance from A equals AB, naming its centre and radius. [2 marks]

1. The locus is a circle with centre A, because the distance is measured from the fixed point A.
2. Its radius is AB, the fixed positive distance specified; points inside the circle are not included.

Q3. A and B are distinct fixed points. Describe a ruler-and-compass construction of the complete locus of P such that PA = PB, where PA and PB are distances. Justify the result. [4 marks]

1. Draw AB and set the compass opening to more than half its length so that equal arcs from its endpoints can intersect.
2. Draw arcs from A and B with this unchanged opening, obtaining intersections C and D on opposite sides of AB.
3. Draw and extend the straight line CD. This is the perpendicular bisector of the given segment AB.
4. Every point of that line is equidistant from A and B, and every point satisfying PA = PB lies on it.

Q4. Two complete straight lines intersect, and a point is unrestricted in their plane. State its locus if it is equidistant from the lines. Explain how the distances are measured and how the answer changes if the point must lie within one specified angle. [3 marks]

1. The unrestricted locus consists of both angle-bisector lines through the intersection, accounting for the angles formed by the given lines.
2. Distance to each given line is the length of the perpendicular from the point to that line, not an arbitrary slanting segment.
3. With the point restricted to one specified angle, retain the bisector ray lying in that angle as the relevant part of the locus.

Q5. A and B are distinct points in a plane. Describe the possible centres of circles through both points, explain how to draw such a circle, and state the least possible radius. [3 marks]

1. The centres form the perpendicular bisector of AB because each centre must be equidistant from the two given points.
2. Choose any point O on this line and draw a circle with centre O and radius OA. Since OA = OB, it passes through A and B.
3. The least radius is AB/2, obtained when O is the midpoint of AB, making the joining segment a diameter.

Q6. Construct triangle ABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Describe how to construct its circumcircle, justify the radius, and state whether the centre lies inside or outside the triangle. [5 marks]

1. Draw the base AB of length 5 cm. At A and B, draw rays making interior angles of 70° and 60° on the same side of AB.
2. Call the intersection of these rays C and complete triangle ABC. Its remaining angle is 180° − 70° − 60° = 50°, so all its angles are acute.
3. Construct the perpendicular bisectors of AB and AC using equal intersecting arcs from each segment's endpoints. Call the intersection of these bisectors O.
4. The bisectors give OA = OB and OA = OC. Draw the circle with centre O and radius OA; these equalities ensure that B and C also lie on it.
5. The centre O lies inside the triangle because the constructed triangle is acute-angled.

Q7. Construct triangle ABC with AB = 5 cm, AC = 4 cm and ∠A = 100°. Explain how to draw its circumcircle and state the centre's position relative to the triangle. [5 marks]

1. Draw AB of length 5 cm. At A, draw a ray forming the given angle of 100° with AB.
2. On this ray mark C at a distance of 4 cm from A. Join BC to complete the triangle determined by the supplied sides and included angle.
3. Construct the perpendicular bisectors of AB and AC, extending them until they meet at a point O outside the triangle.
4. The first bisector gives OA = OB and the second gives OA = OC. Hence the circle with centre O and radius OA passes through all three vertices.
5. The centre lies outside because the angle at A is obtuse; its position agrees with the circumcentre property for an obtuse-angled triangle.

Q8. Triangle ABC has AB = 6 cm, BC = 7 cm and CA = 7 cm. Describe its construction, locate its circumcentre O, and explain the expected relationship between measured lengths OA, OB and OC. [5 marks]

1. Draw the base AB of length 6 cm. With A as centre, draw an arc of radius 7 cm.
2. With B as centre, draw another arc of radius 7 cm. Choose an intersection as C and join AC and BC, obtaining the three prescribed sides.
3. Construct the perpendicular bisectors of AB and AC and label their intersection O. This point is the circumcentre.
4. Draw the circle with centre O and radius OA. The first bisector ensures OA = OB; the second ensures OA = OC, so the circle passes through every vertex.
5. Measure the three radii using the same length unit. They should agree within drawing accuracy, with the exact theoretical relation OA = OB = OC.

Key takeaways

• A locus includes all points satisfying its stated condition, so a single suitable point may be an incomplete answer.
• A fixed positive distance from a fixed point gives a circle with that centre and that distance as radius.
• The locus equidistant from two distinct points is the complete perpendicular bisector of their joining segment.
• For two intersecting lines, compare perpendicular distances and include both angle-bisector lines unless the position is restricted.
• Construct each locus separately when a point must satisfy multiple conditions, then identify and check their common points.
• The centres of circles through two given points lie on their perpendicular bisector; the least radius is half their separation.
• Two perpendicular bisectors locate the circumcentre of a triangle, giving equal distances to all three vertices.
• An obtuse-angled triangle has its circumcentre outside, while an acute-angled triangle has its circumcentre inside.

Test yourself

What must you check before claiming that a proposed shape is the complete locus?

Every point on it must satisfy the condition, and every point satisfying the condition must lie on it.

Does equal distance from two fixed points imply one fixed numerical distance?

No. The two distances agree at each position, but their common value can change along the perpendicular bisector.

Why is the interior excluded from a circle defined by a fixed-distance condition?

Interior points are closer to the centre than the specified radius, so they do not satisfy the equality.

How is the distance from a point to a line measured?

Measure the perpendicular segment from the point to its foot on the line.

What is missing if only one angle-bisector ray is drawn for two unrestricted intersecting lines?

The remaining parts of the complete locus are missing; both full angle-bisector lines are required.

Why does the intersection of two perpendicular bisectors locate a triangle's circumcentre?

One bisector gives equality of distances to one pair of vertices, and the other supplies equality with the third vertex.

Can a circle pass through three distinct collinear points?

No. Three distinct points on one straight line cannot all lie on the same circle.

Where is the circumcentre of a right-angled triangle?

It is at the midpoint of the hypotenuse, the side opposite the right angle.