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IGCSE · Practice questions · Two circles

Sum, difference, subtract.

Six original Cambridge-style questions on two circles: comparing the distance between centres with the sum and difference of the radii, finding the common chord and the points of intersection, points of contact and touching circles, in exact surd form without a calculator.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Centres, radii, distance, then compare.

01
Recall
[2 marks]

The circles \(x^{2} + y^{2} = 16\) and \((x - 6)^{2} + (y - 8)^{2} = 36\) are drawn. State whether they are separate, touch or intersect, giving a reason.

  • Centres \((0, 0)\) and \((6, 8)\), radii 4 and 6; \(d = \sqrt{36 + 64} = 10\). ✓
  • \(d = 10 = 4 + 6 = r_1 + r_2\), so the circles touch externally. ✓
02
Calculation
[3 marks]

Determine whether the circles \(x^{2} + y^{2} + 4x - 2y - 4 = 0\) and \(x^{2} + y^{2} - 8x - 6y + 21 = 0\) intersect.

  • First circle: centre \((-2, 1)\), radius \(\sqrt{4 + 1 + 4} = 3\). Second: centre \((4, 3)\), radius \(\sqrt{16 + 9 - 21} = 2\). ✓
  • \(d = \sqrt{6^{2} + 2^{2}} = \sqrt{40} = 2\sqrt{10}\), and \(r_1 + r_2 = 5\). ✓
  • \(d^{2} = 40 > 25 = 5^{2}\), so \(d > r_1 + r_2\): the circles are separate and do not intersect. ✓
03
Calculation
[5 marks]

The circles \(C_1: x^{2} + y^{2} - 2x - 4y - 20 = 0\) and \(C_2: x^{2} + y^{2} - 12x - 14y + 60 = 0\) meet at \(P\) and \(Q\).

(a) Find the equation of the line \(PQ\).

(b) Find the coordinates of \(P\) and \(Q\).

(c) Find the exact length of \(PQ\).

  • (a) \(C_1 - C_2\): \(10x + 10y - 80 = 0\), so \(PQ\) is \(x + y = 8\). ✓
  • (b) Substitute \(y = 8 - x\) into \(C_1\): \(x^{2} + (8 - x)^{2} - 2x - 4(8 - x) - 20 = 0\). ✓
  • \(2x^{2} - 14x + 12 = 0\), so \(x^{2} - 7x + 6 = 0\) and \((x - 1)(x - 6) = 0\). ✓
  • \(P(1, 7)\) and \(Q(6, 2)\). ✓
  • (c) \(PQ = \sqrt{5^{2} + 5^{2}} = \sqrt{50} = 5\sqrt{2}\). ✓
04
Calculation
[4 marks]

The circle \(C\) has equation \((x - 5)^{2} + (y - 12)^{2} = r^{2}\), where \(r > 0\). Find the two values of \(r\) for which \(C\) touches the circle \(x^{2} + y^{2} = 4\), and say how the circles touch in each case.

  • Centres \((0, 0)\) and \((5, 12)\): \(d = \sqrt{25 + 144} = 13\). The radii are \(2\) and \(r\). ✓
  • Externally: \(r + 2 = 13\), so \(r = 11\). ✓
  • Internally: \(|r - 2| = 13\). \(2 - r = 13\) gives \(r = -11\), rejected, so \(r - 2 = 13\) and \(r = 15\). ✓
  • \(r = 11\): touch externally; \(r = 15\): the small circle touches \(C\) from the inside. ✓
05
Analysis
[5 marks]

The circles \(C_1: x^{2} + y^{2} = 4\) and \(C_2: x^{2} + y^{2} - 6x - 8y + 16 = 0\) are given.

(a) Show that the circles touch externally.

(b) Find the coordinates of the point of contact.

(c) Find the equation of the common tangent at the point of contact.

  • (a) \(C_2\): centre \((3, 4)\), radius \(\sqrt{9 + 16 - 16} = 3\). \(C_1\): centre \((0, 0)\), radius 2. \(d = 5 = 2 + 3\), so they touch externally. ✓
  • (b) The point is \(\frac{2}{5}\) of the way from \((0, 0)\) to \((3, 4)\): \(\left(\frac{6}{5}, \frac{8}{5}\right)\). ✓
  • (c) \(C_1 - C_2\): \((x^{2} + y^{2} - 4) - (x^{2} + y^{2} - 6x - 8y + 16) = 6x + 8y - 20 = 0\), so \(3x + 4y = 10\). ✓
  • Check: \(y = \frac{10 - 3x}{4}\) in \(C_1\) gives \(25x^{2} - 60x + 36 = 0\), i.e. \((5x - 6)^{2} = 0\): a repeated root at \(x = \frac{6}{5}\), so the line is a tangent there. ✓
06
Analysis
[6 marks]

A circle \(C\) has its centre in the first quadrant, touches both coordinate axes and touches the circle \(x^{2} + y^{2} = 1\) externally.

(a) Explain why the centre of \(C\) is \((r, r)\), where \(r\) is its radius, and show that \(r\sqrt{2} = r + 1\).

(b) Find \(r\) in the form \(a + \sqrt{b}\).

(c) Find the equation of \(C\), expanding any brackets in the constants.

(d) Find the exact coordinates of the point of contact of the two circles.

  • (a) Touching the \(x\)-axis puts the centre \(r\) above it, and touching the \(y\)-axis puts it \(r\) to the right, so the centre is \((r, r)\). ✓
  • Its distance from \(O\) is \(\sqrt{r^{2} + r^{2}} = r\sqrt{2}\); touching externally means this equals \(r + 1\). ✓
  • (b) \(r(\sqrt{2} - 1) = 1\), so \(r = \dfrac{1}{\sqrt{2} - 1} = \dfrac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = 1 + \sqrt{2}\). ✓
  • (c) \(r^{2} = (1 + \sqrt{2})^{2} = 3 + 2\sqrt{2}\), so \(C\) is \(\left(x - 1 - \sqrt{2}\right)^{2} + \left(y - 1 - \sqrt{2}\right)^{2} = 3 + 2\sqrt{2}\). ✓
  • (d) The contact point is on the line \(y = x\) at distance 1 from \(O\): \(x\sqrt{2} = 1\), so \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.