Standard forms of a circle
Every point on the rim of a bicycle wheel is at the same distance from the middle. The middle is the centre and that same distance is the radius. Building on classes 9–10 : In classes 9–10 we learnt diameter, chord and radius of a circle; now we write the same circle as an equation using coordinates.
The locus of all points at the same distance (radius) from a fixed point (centre) is a circle. Once the centre (h, k) and the radius r are given, the equation of the circle is completely fixed.
Rule 1: Centre-radius form — (x − h)² + (y − k)² = r². If the centre is the origin it is x² + y² = r².
In the figure : Any point P(x, y) on the circle is at distance r from the centre C(h, k).
Rule 2: From the ends of a diameter — If the ends of a diameter are (x₁, y₁) and (x₂, y₂) then (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0. Or centre = midpoint and r = diameter ÷ 2.
In the figure : A and B are the ends of a diameter; C is the midpoint of AB.
How is the formula obtained? : Take the centre C(h, k) and a point P(x, y) on the circle. CP = r. By the distance formula √((x − h)² + (y − k)²) = r. Squaring both sides gives (x − h)² + (y − k)² = r². For the ends of a diameter: joining P on the circle to the ends A and B gives angle APB = 90°, so the product of the slopes of PA and PB is −1. Writing this and simplifying gives (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0.
- Easy : Centre (−2, 3), radius 4: (x + 2)² + (y − 3)² = 16.
- Medium : Circle with centre (1, −1) passing through (4, 3): r² = (4 − 1)² + (3 + 1)² = 9 + 16 = 25. Equation (x − 1)² + (y + 1)² = 25.
- Exam-type : Ends of a diameter (−1, 2) and (5, −4): (x + 1)(x − 5) + (y − 2)(y + 4) = 0, i.e. x² − 4x − 5 + y² + 2y − 8 = 0, so x² + y² − 4x + 2y − 13 = 0. (The centre (2, −1) and r² = 18 also follow.)
With centre (h, k) the equation contains x − h and y − k; so for the centre (−2, 3) the equation has (x + 2) and (y − 3).
Writing r instead of r² (for example = 4 when the radius is 4). The right side is always the square of the radius.
A circle touching an axis: the distance of the centre from that axis = radius. If it touches both axes the centre is (±r, ±r). Memory clue : "touching means distance = radius".
CET gives the point of contact on an axis; for example touching the X-axis at (3, 0) means the centre is (3, k) and r = |k|. Then put the other given point in the equation.
CET-style example : A circle touches the X-axis at (3, 0) and passes through (1, 2). Its equation is? (A) (x − 3)² + (y − 2)² = 4 (B) (x − 3)² + (y − 1)² = 1 (C) (x − 3)² + (y − 4)² = 16 (D) (x − 1)² + (y − 3)² = 4
Answer : (A). The centre is (3, k) with r = k. (1 − 3)² + (2 − k)² = k², 4 + 4 − 4k = 0, k = 2. The equation is (x − 3)² + (y − 2)² = 4. Time-saving method : Put (1, 2) in the options: in A, 4 + 0 = 4 ✓ and the centre's y equals r (2 = 2) ✓; the others fail one condition.
Spot the difference
| Point | Centre-radius form | Diameter-ends form |
|---|---|---|
| Given | centre and radius | two ends of a diameter |
| Formula | (x − h)² + (y − k)² = r² | (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0 |
| Next step | the equation directly | expand to get the general form |
Practice questions
- 1Write the equation of the circle with centre at the origin and radius 9.
- 2Write the equation of the circle with centre (3, −4) and radius 2.
- 3State the centre and the radius of (x + 5)² + (y − 1)² = 36.
- 4Find the circle with centre (2, 3) passing through (6, 6).
- 5Find the centre and r² of the circle with diameter ends (1, 1) and (5, 7).
- 6A circle of radius 5 in the second quadrant touches both axes. Write its equation.
- 7A circle touches the Y-axis at (0, 4) and passes through (8, 0). Find its equation.
- 8Write the equation of the circle with circumference 10π and centre (−1, −2).
Show answers (try them yourself first)
- 1x² + y² = 81.
- 2(x − 3)² + (y + 4)² = 4.
- 3Centre (−5, 1), radius 6.
- 4r² = (6 − 2)² + (6 − 3)² = 16 + 9 = 25; (x − 2)² + (y − 3)² = 25.
- 5Centre (3, 4); diameter² = 16 + 36 = 52, so r² = 13.
- 6Centre (−5, 5); (x + 5)² + (y − 5)² = 25.
- 7The centre is (r, 4). (8 − r)² + 16 = r², 80 − 16r = 0, r = 5. The equation is (x − 5)² + (y − 4)² = 25.
- 82πr = 10π, r = 5; (x + 1)² + (y + 2)² = 25.
General equation of a circle
When the equation of a fence at equal distance around a village on a map is written with the brackets opened and all terms on one side, we get the general form. Building on classes 9–10 : In classes 9–10 we learnt the expansion of (a + b)² and completing the square; here we use them again and again.
The general form is x² + y² + 2gx + 2fy + c = 0. The multipliers of x² and y² are equal (1) and there is no xy term. For it to be a real circle we need g² + f² − c > 0.
Rule 1: Centre and radius — Centre (−g, −f) and radius r = √(g² + f² − c).
In the figure : x² + y² − 6x + 4y − 12 = 0: g = −3, f = 2, centre (3, −2), r = √(9 + 4 + 12) = 5.
Rule 2: Position of a point and three points — S₁ = x₁² + y₁² + 2gx₁ + 2fy₁ + c: S₁ < 0 inside, S₁ = 0 on, S₁ > 0 outside. If three points are given, put each in x² + y² + 2gx + 2fy + c = 0 and solve the three equations.
How is the formula obtained? : Expand (x − h)² + (y − k)² = r²: x² − 2hx + h² + y² − 2ky + k² − r² = 0. Compare with x² + y² + 2gx + 2fy + c = 0: 2g = −2h, 2f = −2k, c = h² + k² − r². So h = −g, k = −f and r² = h² + k² − c = g² + f² − c.
- Easy : x² + y² − 8x + 6y = 0: 2g = −8, 2f = 6, c = 0. g = −4, f = 3. Centre (4, −3) and r = √(16 + 9 − 0) = 5.
- Medium : Circle through (0, 0), (8, 0) and (0, 6): (0, 0) gives c = 0. (8, 0) gives 64 + 16g = 0, g = −4. (0, 6) gives 36 + 12f = 0, f = −3. The equation is x² + y² − 8x − 6y = 0.
- Exam-type : If the radius of x² + y² + 2x − 6y + k = 0 is 2, find k: g = 1, f = −3. r² = 1 + 9 − k = 4, so k = 6. (For a real circle 10 − k > 0, i.e. k < 10.)
In the general form 2g is the multiplier of x. If the multiplier of x is −8 then g = −4 and the x of the centre is +4.
Taking g = −8 when the multiplier of x is −8. Always halve it and reverse the sign to get the centre.
Centre: half the multipliers of x and y with the sign reversed. r²: sum of the squares of the halved multipliers − c. Memory clue : "halve it, flip the sign".
In CET the multipliers of x² and y² may not be 1 (for example 2x² + 2y² ...); first divide every term to make them 1 and only then read g and f.
CET-style example : Where is the point (2, 1) with respect to the circle x² + y² − 10x − 4y + 4 = 0? (A) inside (B) on (C) outside (D) at the centre
Answer : (A). S₁ = 4 + 1 − 20 − 4 + 4 = −15 < 0, so it is inside. Time-saving method : Only look at the sign: −15 is negative, so inside. The centre is (5, 2); (2, 1) is not the centre.
Spot the difference
| Point | Centre-radius form | General form |
|---|---|---|
| Centre | directly (h, k) | (−g, −f) |
| Radius | r² directly | r = √(g² + f² − c) |
| Useful when | centre and radius are known | three points or an expanded equation are given |
Practice questions
- 1Find the centre and radius of x² + y² + 10x − 4y + 4 = 0.
- 2Write (x − 3)² + (y + 1)² = 16 in the general form.
- 3Is the point (3, −1) inside, on or outside x² + y² − 4x + 2y − 4 = 0?
- 4Find the circle through (0, 0), (6, 0) and (0, −8).
- 5Find the condition on k for x² + y² − 6x + 2y + k = 0 to be a real circle.
- 6If the radius of x² + y² − 4x + 6y + k = 0 is 2, find k.
- 7The circle x² + y² − 4x + 6y + k = 0 touches the Y-axis. Find k.
- 8For the circle with centre (1, 2) and area 49π, find c in x² + y² − 2x − 4y + c = 0.
Show answers (try them yourself first)
- 1g = 5, f = −2; centre (−5, 2) and r = √(25 + 4 − 4) = 5.
- 2x² − 6x + 9 + y² + 2y + 1 − 16 = 0, so x² + y² − 6x + 2y − 6 = 0.
- 3S₁ = 9 + 1 − 12 − 2 − 4 = −8 < 0, so inside.
- 4c = 0; (6, 0): 36 + 12g = 0, g = −3; (0, −8): 64 − 16f = 0, f = 4. The equation is x² + y² − 6x + 8y = 0.
- 5g = −3, f = 1; 9 + 1 − k > 0, so k < 10.
- 6g = −2, f = 3; 4 + 9 − k = 4, so k = 9.
- 7Touching the Y-axis means r = |g|: g² + f² − k = g², so k = f² = 9.
- 8g = −1, f = −2, r² = 49: 1 + 4 − c = 49, so c = −44.
Parametric form of a circle
As the second hand of a clock turns, its tip moves on a circle; when the angle of the hand changes, the place of the tip changes. Here the angle is the changing "parameter". Building on classes 9–10 : In classes 9–10 we learnt sin θ and cos θ; here the same give the coordinates of a point.
If both x and y of a point on the circle are written with one changing number (θ), those equations are the parametric equations of the circle. θ is the angle that the radius from the centre to the point makes with the X-axis.
Rule 1: Parametric equations — Centre (h, k), radius r: x = h + r cos θ, y = k + r sin θ. Centre at the origin: x = r cos θ, y = r sin θ.
In the figure : In the right triangle the horizontal side is r cos θ and the vertical side is r sin θ.
Rule 2: Parametric to Cartesian — Put cos θ = (x − h)/r and sin θ = (y − k)/r and use cos²θ + sin²θ = 1: (x − h)² + (y − k)² = r².
How is the formula obtained? : Join the centre C(h, k) to the point P(x, y) on the circle. Draw a perpendicular from P to the horizontal through C: a right triangle with hypotenuse r. The horizontal side = r cos θ and the vertical side = r sin θ. So x = h + r cos θ and y = k + r sin θ for P.
- Easy : Centre at the origin, r = 5, θ = 90°: (5 cos 90°, 5 sin 90°) = (0, 5).
- Medium : Centre (2, −1), r = 3, θ = 180°: x = 2 + 3 cos 180° = 2 − 3 = −1 and y = −1 + 3 sin 180° = −1. The point is (−1, −1).
- Exam-type : Cartesian equation of x = 1 + 4 cos θ, y = 2 + 4 sin θ: cos θ = (x − 1)/4 and sin θ = (y − 2)/4. Squaring and adding: (x − 1)²/16 + (y − 2)²/16 = 1, so (x − 1)² + (y − 2)² = 16.
For θ = 0°, 90°, 180°, 270° the values of cos and sin are 1, 0, −1; they give the right, top, left and bottom points from the centre.
Pairing cos θ with y and sin θ with x. It is cos with x and sin with y.
Greatest value: take the maximum 1 of cos θ or sin θ. Greatest x = h + r, greatest y = k + r; least x = h − r, least y = k − r. Memory clue : "cos with x, sin with y — both have maximum 1".
CET asks for a combination such as x² + y² − 2x + 6y of the parametric form; (x − h)² + (y − k)² = r² gives the value directly.
CET-style example : The greatest value of y on the circle x = 3 + 5 cos θ, y = −2 + 5 sin θ is? (A) 3 (B) −7 (C) 5 (D) 8
Answer : (A). y = −2 + 5 sin θ and the maximum of sin θ is 1: y = −2 + 5 = 3. Time-saving method : Centre's y (−2) + radius (5) = 3, without any working.
Spot the difference
| Point | Cartesian equation | Parametric equations |
|---|---|---|
| Variables | x and y | only θ |
| Form | (x − h)² + (y − k)² = r² | x = h + r cos θ, y = k + r sin θ |
| Use | testing a point inside or outside | listing points on the circle |
Practice questions
- 1Write the point with θ = 0° on x² + y² = 36.
- 2Write the centre of x = −1 + 4 cos θ, y = 3 + 4 sin θ.
- 3Write the parametric equations of x² + y² = 64.
- 4Write the Cartesian equation of x = −2 + 3 cos θ, y = 4 + 3 sin θ.
- 5Find the point for centre (1, 2), r = 5 and θ = 270°.
- 6Find the point on x² + y² = 100 with θ = 30°.
- 7The point (6, 4) lies on the circle with centre (3, 0) and r = 5. Find cos θ and sin θ for it.
- 8Find the least distance from the origin of the points on x = 3 + 2 cos θ, y = 4 + 2 sin θ.
Show answers (try them yourself first)
- 1(6, 0).
- 2(−1, 3).
- 3x = 8 cos θ, y = 8 sin θ.
- 4(x + 2)² + (y − 4)² = 9.
- 5x = 1 + 5 × 0 = 1, y = 2 + 5 × (−1) = −3. The point is (1, −3).
- 6(10 × √3/2, 10 × 1/2) = (5√3, 5).
- 76 = 3 + 5 cos θ, so cos θ = 3/5. 4 = 5 sin θ, so sin θ = 4/5. (Check: 9/25 + 16/25 = 1.)
- 8The centre (3, 4) is at distance 5 from the origin and r = 2, so the least distance is 5 − 2 = 3.
Tangent to a circle
A bicycle wheel touches a flat road at just one point. The road is a tangent to the wheel and the touching point is the point of contact. Building on classes 9–10 : In classes 9–10 we learnt that "a tangent is perpendicular to the radius"; we now use it to write equations.
A line that meets the circle at exactly one point is a tangent. The length of the segment from an external point to the point of contact is the length of the tangent.
Rule 1: Tangent at a point — To x² + y² = r² at (x₁, y₁): xx₁ + yy₁ = r². To x² + y² + 2gx + 2fy + c = 0: xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0.
In the figure : The radius CP at the point of contact P and the tangent are perpendicular.
Rule 2: Condition of tangency and tangents by slope — y = mx + c touches x² + y² = r² if c² = r²(1 + m²). For a given slope m the tangents are y = mx ± r√(1 + m²). In general: distance of the centre from the line = r.
Rule 3: Length of the tangent — From an external point (x₁, y₁): length = √S₁, where S₁ = x₁² + y₁² + 2gx₁ + 2fy₁ + c. Two tangents of equal length can be drawn from an external point.
In the figure : The two tangents PA and PB from the external point P are equal in length.
How is the formula obtained? : Take the centre at the origin and the point of contact P(x₁, y₁). The radius OP has slope y₁/x₁. The tangent is perpendicular to it, so its slope is −x₁/y₁. y − y₁ = −(x₁/y₁)(x − x₁), i.e. xx₁ + yy₁ = x₁² + y₁² = r². For the length: external point P, centre C, point of contact A. In triangle CAP the angle at A is a right angle, so PA² = CP² − CA² = CP² − r². This value is S₁.
- Easy : Tangent to x² + y² = 100 at (6, 8): 6x + 8y = 100, i.e. 3x + 4y − 50 = 0.
- Medium : Tangents of slope −1 to x² + y² = 50: c² = 50 × (1 + 1) = 100, c = ±10. The lines are y = −x ± 10, i.e. x + y − 10 = 0 and x + y + 10 = 0.
- Exam-type : Length of the tangent from (4, 9) to x² + y² − 2x − 6y + 1 = 0: S₁ = 16 + 81 − 8 − 54 + 1 = 36. Length = √36 = 6.
A tangent at a point is for a point on the circle; two tangents come from an external point, one from a point on the circle and none from an inside point.
Forgetting to check that (x₁, y₁) lies on the circle before writing the tangent. First put the point in the equation.
For xx₁ + yy₁ = r²: "replace x² by x·x₁ and y² by y·y₁". In the general form x² → xx₁, y² → yy₁, 2gx → g(x + x₁), 2fy → f(y + y₁). Memory clue : "split each square, share each half".
When CET says "the line touches", k has two values (±). Taking c² in the condition gives both signs, so check the options.
CET-style example : The line 4x − 3y + k = 0 touches x² + y² = 9. The positive value of k is? (A) 15 (B) 9 (C) 5 (D) 3
Answer : (A). Distance of the centre = r: |k|/√(16 + 9) = 3, |k| = 15, k = 15. Time-saving method : When the origin is the centre, k = r × √(a² + b²) = 3 × 5 = 15.
Spot the difference
| Position of the point | Tangents | Length |
|---|---|---|
| On the circle | one | 0 |
| Outside the circle | two | √S₁ |
| Inside the circle | none | not defined |
Practice questions
- 1Write the tangent to x² + y² = 16 at (0, −4).
- 2Write the tangent to x² + y² = 100 at (8, 6).
- 3Find the length of the tangent from (10, 0) to x² + y² = 64.
- 4Write the tangent to (x − 1)² + (y + 1)² = 16 at (1, 3).
- 5Find the tangents of slope 2 to x² + y² = 20.
- 6The line 3x + 4y + k = 0 touches x² + y² = 16. Find the values of k.
- 7Find the tangent to x² + y² − 2x − 2y − 23 = 0 at (4, 5).
- 8Find the length of the tangent from the origin to x² + y² − 6x − 8y + 9 = 0.
Show answers (try them yourself first)
- 1y = −4, i.e. y + 4 = 0.
- 28x + 6y = 100, i.e. 4x + 3y − 50 = 0.
- 3√(100 − 64) = √36 = 6.
- 4The centre is (1, −1) and the point of contact (1, 3): the radius is vertical, so the tangent is horizontal: y = 3, i.e. y − 3 = 0.
- 5c² = 20 × (1 + 4) = 100, c = ±10. The lines are y = 2x + 10 and y = 2x − 10.
- 6|k|/5 = 4, |k| = 20, so k = 20 or k = −20.
- 7The centre is (1, 1) and the radius at (4, 5) has direction (3, 4). The tangent is 3(x − 4) + 4(y − 5) = 0, so 3x + 4y − 32 = 0.
- 8S₁ = 0 + 0 − 0 − 0 + 9 = 9. Length = 3.
Director circle and line–circle position
Two walls meet at a right angle in a corner; if a round ball touches both walls its centre is at a fixed distance from the corner. A fixed distance of this kind fixes the radius of the director circle. Building on classes 9–10 : In classes 9–10 we learnt Pythagoras and the diagonal of a square; the radius of the director circle comes from them.
The locus of all points from which two tangents to a circle are perpendicular to each other is the director circle of that circle. The position of a line and a circle is decided by comparing the distance d of the centre from the line with the radius r.
Rule 1: Director circle — The director circle of x² + y² = r² is x² + y² = 2r². For (x − h)² + (y − k)² = r² it is (x − h)² + (y − k)² = 2r². Same centre, radius r√2.
In the figure : The two tangents from P on the green director circle are perpendicular.
Rule 2: Line and circle — The distance d of the centre from the line: d < r cuts, d = r touches, d > r does not meet. The length of the chord is 2√(r² − d²).
In the figure : A line at distance d from the centre makes the chord AB; AB = 2√(r² − d²).
Rule 3: Position of two circles — Distance between the centres d, radii r₁ and r₂: d = r₁ + r₂ touch externally; d = |r₁ − r₂| touch internally; d > r₁ + r₂ separate; |r₁ − r₂| < d < r₁ + r₂ intersect at two points.
How is the formula obtained? : Let P be on the director circle with two perpendicular tangents. The points of contact are A and B and the centre is C. In quadrilateral CAPB three angles are 90°, so it is a square with side r. The diagonal of the square CP = r√2. So P is at distance r√2 from the centre and the locus is x² + y² = 2r². For the chord: the perpendicular from the centre bisects the chord; in the right triangle (half chord)² = r² − d².
- Easy : Director circle of x² + y² = 49: x² + y² = 2 × 49 = 98.
- Medium : 5x + 12y − 39 = 0 and x² + y² = 9: d = 39/√(25 + 144) = 39/13 = 3 = r. So the line touches the circle.
- Exam-type : Length of the chord cut from x² + y² = 169 by 5x + 12y − 65 = 0: d = 65/13 = 5. Chord = 2√(169 − 25) = 2 × 12 = 24.
The director circle and the original circle are concentric; only the radius is √2 times.
Writing twice r (2r) instead of 2r² in the director circle. The right side is 2r², not 2r.
In every chord problem draw the right triangle: hypotenuse r, one side d, the other half the chord. Recognise 5-12-13 and 3-4-5. Memory clue : "half chord, d and r = a right triangle".
When CET gives two circles, first bring both to centre-radius form and then compare d with r₁ + r₂ and |r₁ − r₂|.
CET-style example : The circles x² + y² = 16 and (x − 6)² + y² = 4 are? (A) touch externally (B) touch internally (C) intersect at two points (D) separate
Answer : (A). The centres are (0, 0) and (6, 0), d = 6. r₁ = 4, r₂ = 2. r₁ + r₂ = 6 = d, so they touch externally. Time-saving method : Check r₁ + r₂ first; if it equals d the answer is ready.
Spot the difference
| Condition | Line and circle | Two circles |
|---|---|---|
| Touch | d = r | d = r₁ + r₂ or d = absolute value of r₁ − r₂ |
| Intersect | d < r | d is greater than the absolute value of r₁ − r₂ and less than r₁ + r₂ |
| Do not meet | d > r | d > r₁ + r₂ or d is less than the absolute value of r₁ − r₂ |
Practice questions
- 1Write the director circle of x² + y² = 25.
- 2State the centre of the director circle of (x + 1)² + (y − 3)² = 4.
- 3Does 3x + 4y − 10 = 0 cut, touch or miss x² + y² = 4?
- 4Find the length of the chord cut from x² + y² = 29 by 4x + 3y − 10 = 0.
- 5Two circles of radii 7 and 3 have their centres 4 apart. What is their position?
- 6Find the director circle of x² + y² + 6x − 4y − 12 = 0.
- 7The tangents from (a, a) (a > 0) to x² + y² = 25 are perpendicular. Find a.
- 8What length of chord does the circle x² + y² − 6x − 8y = 0 cut on the X-axis?
Show answers (try them yourself first)
- 1x² + y² = 50.
- 2(−1, 3).
- 3d = 10/5 = 2 = r, so the line touches the circle.
- 4d = 10/5 = 2. Chord = 2√(29 − 4) = 2 × 5 = 10.
- 5d = 4 = 7 − 3 = |r₁ − r₂|, so they touch internally.
- 6g = 3, f = −2, r² = 9 + 4 + 12 = 25. The director circle is (x + 3)² + (y − 2)² = 50.
- 7a² + a² = 2 × 25, so a² = 25 and a = 5.
- 8Put y = 0: x² − 6x = 0, x = 0 or 6. The chord = 6.
Lesson 6: Circle — quick revision
| Topic | Formula or rule |
|---|---|
| Centre-radius form | (x − h)² + (y − k)² = r² |
| Diameter-ends form | (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0 |
| General form | x² + y² + 2gx + 2fy + c = 0 |
| Centre and radius | (−g, −f) and √(g² + f² − c) |
| Parametric form | x = h + r cos θ, y = k + r sin θ |
| Tangent (centre at origin) | xx₁ + yy₁ = r² |
| Condition of tangency | c² = r²(1 + m²) |
| Length of tangent | √S₁ |
| Director circle | x² + y² = 2r² |
| Length of chord | 2√(r² − d²) |
🎯 Most important points for the exam
- The right side is r² (the square of the radius).
- In the general form half the multiplier with the sign reversed = the centre.
- Touching an axis: distance of the centre = r.
- A tangent is perpendicular to the radius.
- Sign of S₁: negative inside, zero on, positive outside.
- Right triangle: r, d and half the chord.
Find the centre, find the radius, then follow the condition. Every time check by putting a given point in the equation.