Circles | ISC Class 11 Maths Notes

This note covers the definition of a circle, standard, general, diameter and parametric equations, finding centres and radii, circles through specified points, axis intercepts, the position of a point, and the relative positions and contact of two circles.

What is a circle, and how is its standard equation derived?

Definition: A circle consists of the points in a plane at a fixed distance from one fixed point. The fixed point is its centre; the fixed positive distance is its radius.

A locus is a set of points satisfying a stated condition. For a circle, that condition is constant distance from the centre. Points enclosed by the circle do not belong to this locus: they have smaller distances from the centre.

Use perpendicular coordinate axes, the horizontal x-axis and vertical y-axis. Their intersection is the origin, with coordinates (0, 0). An ordered pair gives a point's horizontal and vertical coordinates, in that order.

Result: the standard equation

Let C(h, k) denote the centre, where h and k are its coordinates. Let r denote the radius, with r > 0, and P(x, y) any point on the circle. Here x and y are the variable coordinates of P.

1. The defining distance condition is CP = r, where CP denotes the length from C to P.
2. The horizontal and vertical coordinate differences are x − h and y − k.
3. The distance formula gives CP² = (x − h)² + (y − k)².
4. Replacing CP² by r² gives the required equation.

(x − h)² + (y − k)² = r². Conversely, any real point satisfying this equation has distance r from C and therefore belongs to the circle. This establishes both directions of the connection between the geometry and its equation.

The drawing shows coordinate axes meeting at O, a circle with centre C labelled (h, k), and a segment from C to a point P labelled (x, y) on the circle. O labels the origin.

See Fig. 10.12 in your NCERT textbook

When the centre is the origin, h = k = 0, so the equation becomes x² + y² = r². Keep the distinction between radius and radius squared: the constant on the right is r², while the radius itself is its positive square root.

How do the centre and radius determine a circle?

A specified centre and positive radius determine the circle directly. Substitute the centre coordinates into the standard equation without changing their signs prematurely. The subtraction already present in the formula determines the signs inside the brackets.

How should signs and squares be handled?

For a negative horizontal centre coordinate, subtracting that coordinate produces a plus sign inside the first bracket. The same rule applies to the vertical coordinate. Square the given radius before placing it on the right side; do not insert the radius itself there.

Worked example 1. Find the equation of the circle with centre (−3, 2) and radius 4.

Answer: Here h = −3, k = 2 and r = 4. Substitution gives (x − (−3))² + (y − 2)² = 4², hence (x + 3)² + (y − 2)² = 16.

The centre lies at x = −3 and y = 2, because these values make both squared expressions zero. That does not make the centre a point on the circle: zero is different from the positive right side. The equation describes points at distance 4 from that centre.

Worked example 2. Find the equation of the circle with centre (2, 2) passing through (4, 5).

Answer: The radius is the distance from the centre to the given point. Thus r² = (4 − 2)² + (5 − 2)² = 4 + 9 = 13, so r = √13. The equation is (x − 2)² + (y − 2)² = 13.

A point on a circle supplies the missing radius when the centre is known. Calculating r² directly is convenient because that is what the equation uses. Taking a square root and then squaring it again adds an unnecessary step.

Check the second answer by substituting the given point: its left side equals 13, the right side. Also read the centre back from the brackets. These checks verify the two separate conditions used to construct the circle.

How do we use the general equation to find the centre and radius?

Expanding the standard equation gives x² + y² − 2hx − 2ky + h² + k² − r² = 0. Its squared terms have equal coefficients, and it contains no xy term, meaning no term involving the product of the two coordinates.

Result: the general form and its conditions

Write the general form as x² + y² + 2gx + 2fy + c = 0. Here g, f and c are real constants; 2g and 2f are the coefficients multiplying x and y, and c is the constant term.

Completing the square means adding the quantity needed to turn a quadratic expression into a squared bracket. It transforms this equation into (x + g)² + (y + f)² = g² + f² − c.

Therefore, the centre is (−g, −f) and the radius is √(g² + f² − c), provided g² + f² − c > 0. The sign condition matters when deciding whether an algebraic equation represents an ordinary real circle.

Worked example 3. Find the centre and radius of x² + y² + 8x + 10y − 8 = 0.

Answer: Rearrange to x² + 8x + y² + 10y = 8. Add 16 and 25 to both sides to obtain (x + 4)² + (y + 5)² = 49. The centre is (−4, −5) and the radius is 7.

Why must the squared coefficients be normalised?

Normalising here means dividing the whole equation by the common nonzero coefficient of x² and y². Only after that division may the coefficients be compared with 2g, 2f and c. Dividing selected terms changes the equation.

Worked example 4. Find the centre and radius of 2x² + 2y² − x = 0.

Answer: Divide every term by 2: x² + y² − x/2 = 0. Completing the square gives (x − 1/4)² + y² = 1/16. Thus the centre is (1/4, 0) and the radius is 1/4.

Equal squared coefficients and absence of an xy term identify the algebraic pattern, but the positive-radius test is still needed. Reading only the coefficient pattern would overlook the single-point and empty cases in the table.

What is the diameter form of a circle's equation?

A chord is a segment joining two points on a circle. A diameter is a chord through the centre. Its midpoint is the centre, and its length is twice the radius. These properties give a direct equation when the two diameter endpoints are known.

Result: the equation from diameter endpoints

Let A(x₁, y₁) and B(x₂, y₂) be distinct endpoints of a diameter. The subscripts 1 and 2 distinguish the coordinates of A and B. Then the centre coordinates are h = (x₁ + x₂)/2 and k = (y₁ + y₂)/2.

The distance formula gives r² = [(x₁ − x₂)² + (y₁ − y₂)²]/4. Substituting the midpoint and this squared radius into the standard equation gives an equation expressed entirely in the endpoint coordinates.

After expansion and collection of terms, it becomes (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0. This is the diameter form. The variable point is still P(x, y); the four endpoint coordinates are fixed data.

How can the form be checked?

Expand its horizontal product to x² − (x₁ + x₂)x + x₁x₂, and its vertical product to y² − (y₁ + y₂)y + y₁y₂. The resulting equation has the required equal squared coefficients and no xy term.

Comparison with the general form recovers the midpoint as the centre. Substitution of either endpoint makes the equation zero, so both endpoints lie on the circle. Together with the midpoint and radius calculation, this checks the diameter construction.

Note: Two points known merely to lie on a circle are not necessarily diameter endpoints. Use the diameter form only when the diameter condition is supplied or established. Otherwise, further information is needed to determine the centre.

Keep the coordinate pairings intact: the horizontal product uses x₁ and x₂, while the vertical product uses y₁ and y₂. Mixing a horizontal coordinate with a vertical coordinate destroys the midpoint relationship from which the formula was derived.

How does the parametric form describe points on a circle?

A parameter is an auxiliary variable used to express the coordinates of points on a curve. For a circle, an angle is a convenient parameter because the radius has a fixed length while its direction changes.

How are the coordinates obtained from an angle?

Let θ, read as theta, be the angle measured anticlockwise from the positive horizontal direction through C to CP. The functions cos θ and sin θ are the cosine and sine of this angle, giving horizontal and vertical components of a unit direction.

The horizontal displacement from C to P is r cos θ and the vertical displacement is r sin θ. Adding the centre coordinates gives the parametric equations: x = h + r cos θ and y = k + r sin θ.

One complete traversal is described by 0 ≤ θ < 2π in radians, where π denotes the circle constant and a full turn measures 2π radians. A radian is the angle subtended at the centre by an arc equal in length to the radius.

Identity: eliminating the parameter

The trigonometric identity cos² θ + sin² θ = 1 verifies the equations. The notation cos² θ means (cos θ)², and sin² θ means (sin θ)². Subtract h and k from the respective coordinates, square the results, and add.

This gives (x − h)² + (y − k)² = r²(cos² θ + sin² θ) = r². Thus each parameter value gives a point on the circle. Conversely, the direction of any radius supplies an angle and hence the corresponding point.

For a centre at the origin, the equations reduce to x = r cos θ and y = r sin θ. For a shifted centre, retain both translations h and k. The angle determines direction from the centre, not distance from the origin.

How is a circle found through three non-collinear points?

Collinear points lie on one straight line. Three distinct non-collinear points determine one circle. Each point supplies an equation for the three unknown constants g, f and c in the general form, so the problem becomes a system of simultaneous equations.

How are the three conditions written?

Let the points be A(x₁, y₁), B(x₂, y₂) and D(x₃, y₃), where the subscript 3 identifies the coordinates of the third point D. The name D keeps C available for the centre.

1. Substitute A to obtain x₁² + y₁² + 2gx₁ + 2fy₁ + c = 0.
2. Substitute B to obtain x₂² + y₂² + 2gx₂ + 2fy₂ + c = 0.
3. Substitute D to obtain x₃² + y₃² + 2gx₃ + 2fy₃ + c = 0.
4. Subtract pairs of equations to eliminate c; solve the remaining linear equations for g and f, then recover c.

Although the circle equation contains squared coordinates, the substituted coordinates are known numbers. The resulting equations are linear in the unknown constants: none contains g², f² or a product of unknown constants. That is why ordinary simultaneous-equation methods apply.

Why does non-collinearity matter?

The centre must be equally distant from A and B, so it lies on the perpendicular bisector of AB: the line through the segment's midpoint at a right angle to it. It must also lie on the perpendicular bisector of AD.

For three distinct non-collinear points, these two bisectors intersect at one centre. Its distance from any of the three points gives the same positive radius. For three distinct collinear points, no ordinary circle passes through all three.

Finally, substitute all three original points into the completed equation. This checks each given condition separately. The algebraic calculation and the perpendicular-bisector interpretation describe the same construction, so the geometry also explains why the non-collinearity condition cannot be omitted.

How can other sufficient data determine a circle?

A centre constrained to a line has fewer possible positions than an unrestricted centre. Two given points on the circle add equal-distance conditions. Combining these facts can determine the centre and radius, but the number of solutions must come from the equations.

How do two points and a line constrain the centre?

Worked example 5. Find the circle through (2, −2) and (3, 4) whose centre lies on x + y = 2.

Answer: Write the centre as (h, k). Equal squared distances give (2 − h)² + (−2 − k)² = (3 − h)² + (4 − k)². Expansion and cancellation yield 2h + 12k = 17.

The centre condition is h + k = 2. Solving gives h = 0.7 and k = 1.3. Then r² = (2 − 0.7)² + (−2 − 1.3)² = 12.58. The circle is (x − 0.7)² + (y − 1.3)² = 12.58.

Subtraction removes both the squared centre coordinates and the radius. The resulting equation is linear in h and k. Combine it with the given line equation before calculating r²; doing so avoids carrying an unnecessary unknown through the calculation.

Worked example 6. Find the circle through (4, 1) and (6, 5) whose centre lies on 4x + y = 16.

Answer: For centre (h, k), equate (4 − h)² + (1 − k)² and (6 − h)² + (5 − k)². Simplifying gives h + 2k = 11. The centre condition is 4h + k = 16.

Solving gives h = 3 and k = 4. Thus r² = (4 − 3)² + (1 − 4)² = 10. The required equation is (x − 3)² + (y − 4)² = 10.

Can sufficient data produce more than one circle?

Worked example 7. Find the circles of radius 5 whose centres lie on the x-axis and which pass through (2, 3).

Answer: A centre on the x-axis has coordinates (h, 0). The distance condition is (2 − h)² + 3² = 5², giving (2 − h)² = 16. Hence h = 6 or h = −2.

The two equations are (x − 6)² + y² = 25 and (x + 2)² + y² = 25. Both have radius 5, both centres lie on the x-axis, and both equations are satisfied by (2, 3).

A squared equation may produce two possibilities. Retain both until the original conditions have been checked. Here both work, so selecting just one centre would discard a valid circle. An additional restriction would be needed to select one of them.

How are the intercepts made by a circle on the axes found?

An axis intercept point is a point where a curve meets an axis. When a circle meets an axis at two distinct points, the length of the segment between them is its intercept on that axis. Distinguish this length from the signed coordinates of the endpoints.

How do we calculate the x-axis intercept?

On the x-axis, y = 0. Substitution into the general equation gives x² + 2gx + c = 0, so x = −g ± √(g² − c). The symbol ± means that both addition and subtraction are to be considered.

If g² − c is positive, there are two intercept points and their separation is 2√(g² − c). If it is zero, there is one contact point. If it is negative, there is no real intersection with the x-axis.

What changes for the y-axis?

On the y-axis, x = 0. Thus y² + 2fy + c = 0 and y = −f ± √(f² − c). For a positive value of f² − c, the intercept length is 2√(f² − c).

Worked example 8. Find the equation of a circle through the origin making nonzero directed intercepts a and b on the coordinate axes. Here a is the other x-axis intercept coordinate and b is the other y-axis intercept coordinate.

Answer: The three points are (0, 0), (a, 0) and (0, b). Substitution into the general equation first gives c = 0, then 2g = −a and 2f = −b. Hence x² + y² − ax − by = 0.

The word directed means that a and b retain their coordinate signs. The corresponding segment lengths are |a| and |b|, where vertical bars denote absolute value. The nonzero condition keeps these three points distinct and non-collinear.

Complete the squares to check this result: (x − a/2)² + (y − b/2)² = (a² + b²)/4. It has centre (a/2, b/2) and radius √(a² + b²)/2. Setting either coordinate to zero recovers the stated intercept points.

How can we decide whether a point is inside, outside or on a circle?

The equation describes the circle itself, while distance from the centre distinguishes the interior and exterior. Let Q(u, v) be the point being tested, where u and v denote its given coordinates. Keep these fixed coordinates separate from the variables x and y in the equation.

What distance comparison is required?

Calculate CQ² = (u − h)² + (v − k)². If CQ² < r², Q lies inside; if CQ² = r², it lies on the circle; if CQ² > r², it lies outside. Comparing squares avoids square roots of nonnegative distances.

For the normalised general form, define S = u² + v² + 2gu + 2fv + c, the value obtained by substituting Q into its left side. Completing the squares shows S = CQ² − r².

Therefore S < 0 means inside, S = 0 means on, and S > 0 means outside, provided the equation represents a real circle. Apply this sign test to the normalised form with squared coefficients equal to positive one.

Worked example 9. Determine whether (−2.5, 3.5) lies inside, outside or on x² + y² = 25.

Answer: The centre is (0, 0) and r² = 25. The squared distance is (−2.5)² + (3.5)² = 6.25 + 12.25 = 18.5. Since 18.5 < 25, the point lies inside the circle.

A point failing the equality is not automatically outside: it may lie inside instead. The direction of the inequality supplies the missing distinction. Use the centre from the equation when computing distance; distance from the origin works only for a circle centred there.

How are the relative positions and contact of two circles determined?

Let C₁(h₁, k₁) and C₂(h₂, k₂) be the centres of two circles, with positive radii r₁ and r₂ respectively. Each subscript identifies the circle to which a quantity belongs. Let d be the distance between the centres.

The distance formula gives d = √[(h₂ − h₁)² + (k₂ − k₁)²]. Compare d with both r₁ + r₂ and |r₁ − r₂|. The sum controls external separation, while the absolute difference controls containment of one circle within the other.

How do the distance conditions classify the circles?

Concentric circles have the same centre, so d = 0. If their radii differ, they have no common point. If their radii agree, they coincide. The positive-distance restriction in the internal-contact condition prevents coincidence from being misclassified as one-point contact.

Why do sums and differences of radii appear?

At external contact, the contact point lies between the centres, and the two radii along the line of centres add to d. At internal contact, both centres lie on the same side of the contact point, and their distance is the difference of the radii.

Draw two separate sketches with centres C₁ and C₂. Label the contact point T and join it to both centres. In the external sketch put T between the centres; in the internal sketch put the smaller circle inside the larger one.

For two intersection points, joining either common point to the centres forms a triangle. Its side lengths are d, r₁ and r₂. The strict triangle inequalities give |r₁ − r₂| < d < r₁ + r₂; equality corresponds to contact instead.

How is contact used to find an unknown circle?

Write its unknown centre and radius in standard form. Translate any centre restriction into an equation, and translate each given point into a distance equation. Add d = r₁ + r₂ for external contact or d = |r₁ − r₂| > 0 for internal contact.

Solve these equations together. If squaring was used to remove a square root or absolute value, check each solution in the original contact condition. Require positive radii and all stated centre or point restrictions. A contact condition by itself does not usually determine a unique circle.

Glossary

• Circle — The set of points in a plane at a fixed positive distance from one fixed point.
• Centre — The fixed point from which every point on a circle has the same distance.
• Radius — The positive distance from a circle's centre to any point on the circle.
• Locus — The set of all points satisfying a specified geometric condition.
• Chord — A line segment with both its endpoints on a circle.
• Diameter — A chord passing through the centre, with length twice the radius.
• Standard form — The equation (x − h)² + (y − k)² = r², displaying the centre coordinates and squared radius.
• General form — The equation x² + y² + 2gx + 2fy + c = 0, with constants subject to a positive-radius condition.
• Parameter — An auxiliary variable used to express the coordinates of points on a curve.
• Non-collinear points — Points which do not all lie on the same straight line.
• Perpendicular bisector — A line passing through a segment's midpoint at a right angle to that segment.
• Axis intercept — The segment cut off on a coordinate axis, or its length when specified.
• Concentric circles — Circles with the same centre, whose radii may be different.
• Coincident circles — Two descriptions of the same circle, with equal centres and radii.
• Contact — The situation in which two distinct circles share exactly one point.

Common errors and misconceptions

• Misconception: A plus sign in (x + 3)² gives a positive centre coordinate. Correct: The standard form subtracts the centre coordinate, so this bracket corresponds to h = −3.
• Misconception: The right side of the standard equation is the radius. Correct: It is the radius squared; take its positive square root to obtain the radius.
• Misconception: The coefficient of x in the general form is g. Correct: It is 2g. First normalise the equation, then halve the linear coefficients and reverse their signs for the centre.
• Misconception: Equal squared coefficients guarantee an ordinary real circle. Correct: There must also be no xy term and the computed radius squared must be positive.
• Misconception: Any two points on a circle can be used as diameter endpoints. Correct: The diameter condition must be stated or proved before applying diameter form.
• Misconception: Three distinct points determine a circle regardless of their position. Correct: They must be non-collinear; three distinct collinear points cannot lie on one ordinary circle.
• Misconception: To find the x-axis intercept, set x = 0. Correct: Set y = 0, because every point on the x-axis has zero vertical coordinate.
• Misconception: Equal centres and radii mean internal contact. Correct: They mean coincidence; internal contact between distinct circles requires d = |r₁ − r₂| > 0.

Exam-style questions with model answers

Q1. Find the centre and radius of (x + 5)² + (y − 3)² = 36. [2 marks]

1. Compare with (x − h)² + (y − k)² = r²: the centre coordinates are h = −5 and k = 3, so the centre is (−5, 3).
2. The radius r is positive and satisfies r² = 36. Therefore the radius is 6.

Q2. Find the equation of the circle with centre (2, 2) passing through (4, 5). [3 marks]

1. Let r be the radius. Its square is the squared distance between the centre and the given point: r² = (4 − 2)² + (5 − 2)².
2. Evaluate the squares to obtain r² = 4 + 9 = 13. Thus the radius is √13, using the positive square root.
3. Substitute the centre and squared radius into standard form. The required equation is (x − 2)² + (y − 2)² = 13.

Q3. Find the centre and radius of x² + y² + 8x + 10y − 8 = 0 by completing the square. [4 marks]

1. Move the constant to the right and group terms: (x² + 8x) + (y² + 10y) = 8.
2. Add 16 and 25 to both sides, the squares of half the respective linear coefficients: (x² + 8x + 16) + (y² + 10y + 25) = 49.
3. Rewrite the left side as squared brackets: (x + 4)² + (y + 5)² = 49.
4. Comparison with standard form gives centre (−4, −5) and positive radius √49 = 7.

Q4. Find the circle through (2, −2) and (3, 4) whose centre lies on x + y = 2. [5 marks]

1. Let the centre be (h, k), where h and k are its coordinates, and let r be the radius. Since the centre lies on the stated line, h + k = 2.
2. Both given points are at distance r from the centre, so (2 − h)² + (−2 − k)² = r² and (3 − h)² + (4 − k)² = r².
3. Equate these squared distances, expand, and cancel the squared centre coordinates. The resulting linear equation is 2h + 12k = 17.
4. Solving with h + k = 2 gives h = 0.7 and k = 1.3. Substituting the first point gives r² = 1.69 + 10.89 = 12.58.
5. The required circle is (x − 0.7)² + (y − 1.3)² = 12.58. The second point also gives 2.3² + 2.7² = 12.58, verifying its inclusion.

Q5. Find all circles of radius 5 with centres on the x-axis that pass through (2, 3). [4 marks]

1. Write the centre as (h, 0), where h is its horizontal coordinate. The vertical coordinate is zero because the centre lies on the x-axis.
2. The distance from the centre to (2, 3) must be 5, so (2 − h)² + 3² = 25, giving (2 − h)² = 16.
3. Taking both signs gives 2 − h = 4 or 2 − h = −4. Thus h = −2 or h = 6.
4. The required circles are (x + 2)² + y² = 25 and (x − 6)² + y² = 25. Both satisfy every given condition.

Q6. A circle passes through the origin and has other axis intercept points (a, 0) and (0, b), where a and b are nonzero real coordinates. Find its equation, centre and radius. [5 marks]

1. Write the equation as x² + y² + 2gx + 2fy + c = 0, where g, f and c are unknown constants. The origin (0, 0) lies on it, so c = 0.
2. Substitute (a, 0): a² + 2ga = 0. Since a is nonzero, division by a gives 2g = −a.
3. Substitute (0, b): b² + 2fb = 0. Since b is nonzero, division by b gives 2f = −b.
4. The required equation is x² + y² − ax − by = 0. Completing the squares gives (x − a/2)² + (y − b/2)² = (a² + b²)/4.
5. Read off the centre as (a/2, b/2) and the radius as √(a² + b²)/2. The radius is positive because the stipulated intercept coordinates are nonzero.

Q7. Two circles have distinct centres separated by distance d and positive radii r₁ and r₂. State the conditions for external contact, two intersections and internal contact, explaining each. [3 marks]

1. External contact requires d = r₁ + r₂. The contact point is between the centres, so the two radii along their joining line add to the centre distance.
2. Two intersections require |r₁ − r₂| < d < r₁ + r₂. Here |r₁ − r₂| is the absolute difference of the radii; both triangle inequalities are strict.
3. Internal contact requires d = |r₁ − r₂| > 0. The smaller circle touches the larger from within, and the radii differ by the distance between their centres.

Q8. Determine whether (−2.5, 3.5) lies inside, outside or on x² + y² = 25. [2 marks]

1. The centre is the origin. The squared distance of the given point from it is (−2.5)² + 3.5² = 6.25 + 12.25 = 18.5.
2. The radius squared is 25. Since 18.5 < 25, the point lies inside the circle.

Key takeaways

• A circle is defined by a fixed centre and a fixed positive distance, its radius.
• The standard form displays the centre through subtracted coordinates and places the squared radius on the right.
• Normalise the general equation before reading its coefficients, and check that the calculated radius squared is positive.
• Diameter form requires actual diameter endpoints; two arbitrary points on the circle do not supply that condition.
• Parametric equations express both coordinates through one angle and satisfy the standard equation by the sine-cosine identity.
• Three distinct non-collinear points determine one circle; centre restrictions and given points supply further equation-solving methods.
• Find axis intersections by setting the other coordinate to zero, then distinguish endpoint coordinates from intercept lengths.
• Compare centre distance with both the sum and absolute difference of radii when classifying two circles.

Test yourself

What do h, k and r represent in (x − h)² + (y − k)² = r²?

The numbers h and k are the centre coordinates, and r is the positive radius of the circle.

How do you recover the centre from x² + y² + 2gx + 2fy + c = 0?

The centre is (−g, −f). Halve each linear coefficient and reverse its sign after normalising the squared coefficients.

When does the general equation describe an ordinary real circle?

It describes an ordinary real circle when g² + f² − c is positive, giving a positive radius.

Why is non-collinearity required when a circle is determined by three distinct points?

The perpendicular bisectors of two joining segments then meet at one centre. Three distinct collinear points cannot lie on an ordinary circle.

How do you find the intersections of a circle with the x-axis?

Set y = 0 in its equation and solve for x. The real solutions give the horizontal coordinates of the intersection points.

What identity checks x = h + r cos θ and y = k + r sin θ?

Use cos² θ + sin² θ = 1. Squaring the coordinate displacements and adding gives the radius squared.

For distinct circles with radii r₁ and r₂, what distinguishes internal from external contact?

Internal contact has centre distance |r₁ − r₂| > 0; external contact has centre distance r₁ + r₂.

What happens when two circles have the same centre and the same positive radius?

They coincide, representing the same circle and sharing every point, rather than touching at just one point.