Circles ISC Class 11: Forms of the Equation and Two Circles

What Circles is about

Weightage: Circles belongs to the unit Coordinate Geometry, which CISCE weights at 20 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Standard form

If C(h, k) is the centre and r the radius, any point P(x, y) on the circle satisfies CP = r, so (x − h)² + (y − k)² = r². With the centre at the origin it becomes x² + y² = r². Special cases: a circle touching the x-axis has r = |k|; touching the y-axis has r = |h|; touching both axes has |h| = |k| = r. A point (x₁, y₁) lies inside, on or outside the circle according as (x₁ − h)² + (y₁ − k)² is less than, equal to, or greater than r².

Worked example

Find the equation of the circle with centre (2, −3) and radius 5, in general form.

Try it first, then show the solutionSolution

1. (x − 2)² + (y + 3)² = 25
2. x² − 4x + 4 + y² + 6y + 9 − 25 = 0

Answer: x² + y² − 4x + 6y − 12 = 0

General form; centre and radius

Every circle can be written x² + y² + 2gx + 2fy + c = 0. Completing squares: (x + g)² + (y + f)² = g² + f² − c, so the centre is (−g, −f) and the radius is √(g² + f² − c). Features of a circle's equation: it is of second degree, the coefficients of x² and y² are equal, and there is no xy term. If the x² and y² coefficients are equal but not 1, divide through first. If g² + f² − c = 0 the 'circle' is a single point; if it is negative there is no real circle.

Worked example

Find the centre and radius of 2x² + 2y² − 8x + 12y − 6 = 0.

Try it first, then show the solutionSolution

1. Divide by 2: x² + y² − 4x + 6y − 3 = 0, so 2g = −4, 2f = 6, c = −3
2. Centre (−g, −f) = (2, −3)
3. r = √(4 + 9 + 3) = 4

Answer: Centre (2, −3), radius 4

Diameter form

If A(x₁, y₁) and B(x₂, y₂) are the ends of a diameter and P(x, y) is any other point on the circle, then angle APB = 90° (angle in a semicircle), so the slopes of PA and PB multiply to −1. This gives (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0, which is also satisfied by A and B themselves. The centre is the midpoint of AB and the radius is half of AB.

Worked example

Find the circle on the join of (1, 2) and (5, −4) as diameter.

Try it first, then show the solutionSolution

1. (x − 1)(x − 5) + (y − 2)(y + 4) = 0
2. x² − 6x + 5 + y² + 2y − 8 = 0 ⇒ x² + y² − 6x + 2y − 3 = 0
3. Check: centre (3, −1) is the midpoint; r = √(9 + 1 + 3) = √13 = ½√(16 + 36) ✓

Answer: x² + y² − 6x + 2y − 3 = 0

Parametric form

The point (h + r cos θ, k + r sin θ) is at distance r from (h, k) for every θ, so x = h + r cos θ, y = k + r sin θ (0 ≤ θ < 2π) are parametric equations of the circle; θ is the angle the radius to the point makes with the positive x-direction. To convert back, isolate cos θ and sin θ and use cos²θ + sin²θ = 1. Parametric form is convenient for naming a general point on a circle.

Worked example

Find the Cartesian equation of x = 3 + 4 cos θ, y = −1 + 4 sin θ, and the point with θ = π/3.

Try it first, then show the solutionSolution

1. cos θ = (x − 3)/4, sin θ = (y + 1)/4; squaring and adding: (x − 3)² + (y + 1)² = 16
2. θ = π/3: x = 3 + 2 = 5, y = −1 + 2√3

Answer: (x − 3)² + (y + 1)² = 16; (5, −1 + 2√3)

Circle through three non-collinear points

Substitute each point into x² + y² + 2gx + 2fy + c = 0 to get three linear equations in g, f and c. Subtracting pairs of equations eliminates c, leaving two equations for g and f; then find c. (Alternatively, the centre is where the perpendicular bisectors of two chords meet.) Three collinear points give no circle, because the equations are then inconsistent.

Worked example

Find the circle through (1, 1), (2, −1) and (3, 2).

Try it first, then show the solutionSolution

1. 2g + 2f + c = −2; 4g − 2f + c = −5; 6g + 4f + c = −13
2. Subtracting: 2g − 4f = −3 and 2g + 6f = −8 ⇒ 10f = −5, f = −1/2, g = −5/2
3. c = −2 + 5 + 1 = 4; circle x² + y² − 5x − y + 4 = 0
4. Check (3, 2): 9 + 4 − 15 − 2 + 4 = 0 ✓

Answer: x² + y² − 5x − y + 4 = 0; centre (5/2, 1/2), radius √10/2

Circle from other sufficient data

Count the unknowns and use each condition once. Centre on a given line: write the centre with one unknown, e.g. (h, h − 1) for x − y = 1, then use CP₁ = CP₂ for two points on the circle. Touching an axis: the radius equals the distance from the centre to that axis. Touching both axes: centre (a, a) (or with signs for other quadrants) and radius |a|. The centre always lies on the perpendicular bisector of any chord.

Worked example

Find the circle whose centre lies on x − y = 1 and which passes through (2, 3) and (4, 5).

Try it first, then show the solutionSolution

1. Centre (h, h − 1); equal distances: (h − 2)² + (h − 4)² = (h − 4)² + (h − 6)²
2. (h − 2)² = (h − 6)² ⇒ h = 4; centre (4, 3)
3. r² = (4 − 2)² + 0² = 4

Answer: (x − 4)² + (y − 3)² = 4, i.e. x² + y² − 8x − 6y + 21 = 0

Intercepts on the axes

Putting y = 0 in x² + y² + 2gx + 2fy + c = 0 gives x² + 2gx + c = 0, whose roots x₁, x₂ are where the circle meets the x-axis. The length of the x-intercept is |x₁ − x₂| = 2√(g² − c); similarly the y-intercept is 2√(f² − c). If g² − c < 0 the circle does not meet the x-axis; if g² = c it touches the x-axis.

Worked example

Find the intercepts made by x² + y² − 6x − 4y + 5 = 0 on the axes.

Try it first, then show the solutionSolution

1. g = −3, f = −2, c = 5
2. x-intercept 2√(9 − 5) = 4 (it meets the x-axis at x = 1 and x = 5)
3. y-intercept: f² − c = 4 − 5 < 0, so the circle does not meet the y-axis

Answer: 4 on the x-axis; none on the y-axis

Relative position of two circles; touching circles

Let the centres be d apart, with radii r₁ ≥ r₂. d > r₁ + r₂: separate (4 common tangents). d = r₁ + r₂: touch externally (3). r₁ − r₂ < d < r₁ + r₂: intersect in two points (2). d = r₁ − r₂: touch internally (1). d < r₁ − r₂: one inside the other (0). The point of contact lies on the line of centres: it divides C₁C₂ internally in the ratio r₁ : r₂ for external touching, and externally in the ratio r₁ : r₂ for internal touching.

Worked example

Show that x² + y² = 9 and x² + y² − 8x − 6y + 21 = 0 touch, and find the point of contact.

Try it first, then show the solutionSolution

1. C₁(0, 0), r₁ = 3; C₂(4, 3), r₂ = √(16 + 9 − 21) = 2
2. d = 5 = r₁ + r₂, so they touch externally
3. Contact point divides C₁C₂ in 3 : 2: ((3 × 4 + 2 × 0)/5, (3 × 3 + 2 × 0)/5) = (12/5, 9/5)

Answer: Touch externally at (12/5, 9/5)

Maths in real life: Circles

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Circles through Points, Intercepts and Two Circles
• Equation of a Circle: Standard, General and Diameter Forms

Circles formula sheet

Standard form

$(x-h{)}^2+(y-k{)}^2=r^2$

Centre (h, k), radius r.

General form

$x^2+y^2+2gx+2fy+c=0,C(-g,-f),r=\sqrt{g^2+f^2-c}$

Divide so x² and y² have coefficient 1.

Diameter form

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

Angle in a semicircle is 90°.

Parametric form

$x=h+r\cos \theta ,y=k+r\sin \theta$

0 ≤ θ < 2π.

Intercepts

$2\sqrt{g^2-c}(x\text{-axis});2\sqrt{f^2-c}(y\text{-axis})$

Touches the x-axis if g² = c.

Two circles

$d=r_1+r_2\text{ (external touch)};d=|r_1-r_2|\text{ (internal touch)}$

Between these values the circles cut in two points.

Circles: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Circles

The question patterns to expect, and the method that works for each.

Tricks to remember Circles

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Centre: change the signs, halve

  Centre of x² + y² + 2gx + 2fy + c = 0 is (−g, −f): take the x and y coefficients, halve and change sign.

  Use it when: Reading the centre quickly.

• Three equal checks

  Circle equation: equal x², y² coefficients, no xy term, g² + f² − c > 0.

  Use it when: Recognising a circle.

• Sum touches outside, difference touches inside

  d = r₁ + r₂: external touch; d = |r₁ − r₂|: internal touch.

  Use it when: Two circles.

• Intercept = 2√(half-coefficient² − c)

  x-intercept 2√(g² − c), y-intercept 2√(f² − c).

  Use it when: Intercepts on axes.

• Three unknowns, three facts

  g, f, c need three independent conditions.

  Use it when: Planning a 'find the circle' question.

Circles: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Circles

The slips examiners see most often, and what to write instead.

• ✗Mistake: Reading the centre without dividing

  Students read g and f from 2x² + 2y² − 8x + … directly.

  Wrong working

  Centre (4, −6).

  Right working

  Divide by 2 first: centre (2, −3).

  ✓Do this: Make the x² and y² coefficients 1 before reading anything.

  Usually costs: Full marks for the part.

• ✗Mistake: Wrong signs on the centre

  Students take (g, f) instead of (−g, −f).

  Wrong working

  x² + y² + 6x − 4y − 3 = 0 has centre (3, −2).

  Right working

  Centre (−3, 2).

  ✓Do this: Expand (x + 3)² to see the sign.

  Usually costs: 1-2 marks.

• ✗Mistake: Sign of c in the radius

  Students add c instead of subtracting it.

  Wrong working

  x² + y² − 4x + 6y − 3 = 0: r = √(4 + 9 − 3) = √10.

  Right working

  r = √(4 + 9 − (−3)) = 4.

  ✓Do this: Write g² + f² − c with c in brackets.

  Usually costs: 1 mark.

• ✗Mistake: Mixing up internal and external touching

  Students always use d = r₁ + r₂.

  Wrong working

  Circle centre (3, 4) touching x² + y² = 1 internally: r = 4.

  Right working

  Internal: d = |r − 1| = 5 ⇒ r = 6.

  ✓Do this: Draw both cases before writing the condition.

  Usually costs: 1-2 marks.

• ✗Mistake: Forgetting both circles when touching both axes

  The quadratic for a gives two values.

  Wrong working

  Only (x − 1)² + (y − 1)² = 1.

  Right working

  Both a = 1 and a = 5 work.

  ✓Do this: Keep every root that satisfies the conditions.

  Usually costs: 1 mark.

Practice questions on Circles

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The radius of the circle x² + y² − 8x + 6y = 0 is:
   (a) 3
   (b) 4
   (c) 5
   (d) 25

   Hint

   √(g² + f² − c).

   Answer

   (c)

2. Q2 · 1 mark · MCQExam level

   The circle with centre at the origin and radius 3 is:
   (a) x² + y² = 3
   (b) x² + y² = 9
   (c) x² + y² = 6
   (d) (x − 3)² + y² = 9

   Hint

   x² + y² = r².

   Answer

   (b)

3. Q3 · 1 mark · Very short answerExam level

   Write parametric equations of the circle x² + y² = 16.

   Hint

   r = 4.

   Answer

   x = 4 cos θ, y = 4 sin θ

4. Q4 · 1 mark · Very short answerExam level

   Does the point (1, 2) lie inside, on or outside x² + y² = 4?

   Hint

   Compare 1 + 4 with 4.

   Answer

   Outside (1 + 4 = 5 > 4)

5. Q5 · 2 marks · Short answerExam level

   Find the equation of the circle with centre (1, −1) that passes through the origin.

   Hint

   r² = 1 + 1.

   Answer

   (x − 1)² + (y + 1)² = 2 ⇒ x² + y² − 2x + 2y = 0

6. Q6 · 2 marks · Short answerHarder exam level

   Find the centre and radius of x² + y² + 10x − 6y + 9 = 0.

   Hint

   g = 5, f = −3.

   Answer

   Centre (−5, 3); radius √(25 + 9 − 9) = 5

7. Q7 · 2 marks · Short answerHarder exam level

   Find the intercepts made by x² + y² − 5x − 7y + 6 = 0 on the coordinate axes.

   Hint

   g = −5/2, f = −7/2.

   Answer

   x-axis: 2√(25/4 − 6) = 1; y-axis: 2√(49/4 − 6) = 5

8. Q8 · 3 marks · Short answerHarder exam level

   Find the circle drawn on the join of (3, 4) and (−1, 2) as diameter.

   Hint

   Diameter form.

   Answer

   (x − 3)(x + 1) + (y − 4)(y − 2) = 0 ⇒ x² + y² − 2x − 6y + 5 = 0

9. Q9 · 3 marks · Short answerHarder exam level

   Find the circle whose centre lies on 2x + y = 3 and which passes through (−1, 2) and (3, 2).

   Hint

   The centre is on the perpendicular bisector x = 1.

   Answer

   Centre (1, 1); r² = (1 + 1)² + (1 − 2)² = 5; x² + y² − 2x − 2y − 3 = 0

10. Q10 · 3 marks · Short answerHarder exam level

    Show that x² + y² = 16 and x² + y² − 6x + 8y + 24 = 0 touch, and find the point of contact.

    Hint

    Second circle: centre (3, −4), radius 1.

    Answer

    d = 5 = 4 + 1, so they touch externally. Contact divides (0, 0)–(3, −4) in 4 : 1: (12/5, −16/5).

11. Q11 · 2 marks · Short answerHarder exam level

    Find the Cartesian equation of the circle x = −2 + 5 cos θ, y = 3 + 5 sin θ.

    Hint

    cos²θ + sin²θ = 1.

    Answer

    (x + 2)² + (y − 3)² = 25 ⇒ x² + y² + 4x − 6y − 12 = 0

12. Q12 · 3 marks · Short answerHarder exam level

    Find k if the circle x² + y² − 4x + 2y + k = 0 touches the y-axis.

    Hint

    y-intercept zero: f² = c.

    Answer

    f = 1, so k = f² = 1. Check: radius √(4 + 1 − 1) = 2 = distance of centre (2, −1) from the y-axis.

13. Q13 · 5 marks · Long answerHarder exam level

    (a) Show that x² + y² − 2x = 0 and x² + y² + 6x − 6y + 2 = 0 touch each other. [2]
    (b) Find the point of contact. [2]
    (c) How many common tangents do they have? [1]

    Hint

    Centres (1, 0) and (−3, 3).

    Answer

    (a) r₁ = 1, r₂ = √(9 + 9 − 2) = 4, d = 5 = r₁ + r₂: external touching. (b) Divide (1, 0)–(−3, 3) in 1 : 4: ((−3 + 4)/5, (3 + 0)/5) = (1/5, 3/5). (c) 3.

14. Q14 · 3 marks · Short answerHOTS / competency

    Find the circle that touches the x-axis at (3, 0) and passes through (1, 4).

    Hint

    Centre (3, k), radius |k|.

    Answer

    (1 − 3)² + (4 − k)² = k² ⇒ 20 − 8k = 0 ⇒ k = 5/2. (x − 3)² + (y − 5/2)² = 25/4 ⇒ x² + y² − 6x − 5y + 9 = 0

15. Q15 · 3 marks · Short answerHOTS / competency

    Find the circle concentric with x² + y² − 6x + 4y − 3 = 0 whose area is twice the area of this circle.

    Hint

    Area ∝ r².

    Answer

    Centre (3, −2), r² = 9 + 4 + 3 = 16; new r² = 32: (x − 3)² + (y + 2)² = 32 ⇒ x² + y² − 6x + 4y − 19 = 0

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Standard form?Show answer

   (x − h)² + (y − k)² = r².

2. 2.Centre and radius of x² + y² + 2gx + 2fy + c = 0?Show answer

   (−g, −f); √(g² + f² − c).

3. 3.Diameter form?Show answer

   (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0.

4. 4.Parametric form?Show answer

   x = h + r cos θ, y = k + r sin θ.

5. 5.x-intercept length?Show answer

   2√(g² − c).

6. 6.Condition to touch the y-axis?Show answer

   f² = c (radius = |h|).

7. 7.External touching?Show answer

   d = r₁ + r₂.

8. 8.Internal touching?Show answer

   d = |r₁ − r₂|.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write a circle's equation from its centre and radius?
• ☐Can you find the centre and radius when the x² coefficient is not 1?
• ☐Can you write the diameter form and check it?
• ☐Can you convert between parametric and Cartesian forms?
• ☐Can you find the circle through three given points?
• ☐Can you find a circle whose centre lies on a given line?
• ☐Can you find the intercepts on both axes?
• ☐Can you decide whether two circles touch and find the point of contact?

Circles: frequently asked questions