Home / IGCSE Add Maths 0606 / Coordinate geometry of the circle / Two circles
Cambridge IGCSE · Topic 8.4
IGCSE Add Maths 0606 / Topic 8 / IGCSE

Measure the gap between the centres.

Two circles can miss each other, touch, cross at two points or sit one inside the other. One distance decides which: the distance \(d\) between the centres, compared with the sum and the difference of the radii. When they do cross, subtracting one equation from the other hands you the common chord in a single line.

Topic 8All coordinate geometry of the circle lessons →
The key idea

For circles with centres \(A\), \(B\), radii \(r_1\), \(r_2\) and \(d = AB\): \(d > r_1 + r_2\) separate; \(d = r_1 + r_2\) touch externally; \(|r_1 - r_2| < d < r_1 + r_2\) meet at two points; \(d = |r_1 - r_2|\) touch internally; \(d < |r_1 - r_2|\) one inside the other. Subtracting the two equations (each written with \(x^{2} + y^{2}\) on its own) removes \(x^{2}\) and \(y^{2}\) and leaves a straight line: the common chord, or the common tangent if the circles touch.

x y O A(3, 3) B(7, 7) P(0, 7) Q(7, 0) M r₂ = 7 r₁ = 5 circle C₁ circle C₂ common chord line of centres x + y = 7 (C₁ − C₂) AB = 4√2 2 < AB < 12 so two points AB is at right angles to PQ and bisects it at M
Fig. 1: circle C1, centre A(3, 3), radius 5, and circle C2, centre B(7, 7), radius 7. AB = 4√2 lies between 7 - 5 = 2 and 7 + 5 = 12, so the circles meet at two points, P(0, 7) and Q(7, 0). Subtracting the equations gives the common chord x + y = 7. The line of centres y = x is perpendicular to the chord and bisects it at M(3.5, 3.5)
Section 01

Five ways two circles can sit.

Start every two-circle question by finding both centres and both radii (from the general form \(x^{2} + y^{2} + 2gx + 2fy + c = 0\) the centre is \((-g, -f)\) and \(r = \sqrt{g^{2} + f^{2} - c}\)). Then work out the distance between the centres, \(d = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}\), and compare it with the sum \(r_1 + r_2\) and the difference \(|r_1 - r_2|\) of the radii. Picture sliding one circle towards the other: far apart, then touching on the outside, then crossing, then touching on the inside, then lying completely inside.

Distance between centresHow the circles sitCommon points
\(d > r_1 + r_2\)separate, outside each other0
\(d = r_1 + r_2\)touch externally1
\(|r_1 - r_2| < d < r_1 + r_2\)intersect2
\(d = |r_1 - r_2|\)touch internally1
\(d < |r_1 - r_2|\)one inside the other0
Worked example Classify three pairs
Decide how each pair of circles sit. (a) \(x^{2} + y^{2} = 4\) and \(x^{2} + y^{2} - 12x - 16y + 75 = 0\) (b) \(x^{2} + y^{2} - 2x - 4y - 4 = 0\) and \(x^{2} + y^{2} - 8x - 12y + 27 = 0\) (c) \(x^{2} + y^{2} = 100\) and \((x - 3)^{2} + (y - 4)^{2} = 25\)
  1. (a) Centres \((0, 0)\) and \((6, 8)\); radii \(2\) and \(\sqrt{36 + 64 - 75} = 5\). \(d = \sqrt{36 + 64} = 10\) and \(r_1 + r_2 = 7\).
  2. \(10 > 7\), so the circles are separate.
  3. (b) Centres \((1, 2)\) and \((4, 6)\); radii \(\sqrt{1 + 4 + 4} = 3\) and \(\sqrt{16 + 36 - 27} = 5\). \(d = \sqrt{9 + 16} = 5\).
  4. \(|5 - 3| = 2 < 5 < 8 = 5 + 3\), so they meet at two points.
  5. (c) Centres \((0, 0)\) and \((3, 4)\); radii \(10\) and \(5\). \(d = 5 = 10 - 5\), so they touch internally.
Answer: (a) separate   (b) intersect at two points   (c) touch internally
Worked example Find k so the circles touch
Find the values of \(k\) for which the circles \(x^{2} + y^{2} = 9\) and \((x - k)^{2} + (y - 4)^{2} = 4\) touch.
  1. Centres \((0, 0)\) and \((k, 4)\), radii \(3\) and \(2\), so \(d^{2} = k^{2} + 16\).
  2. Touching externally: \(d = 3 + 2 = 5\), so \(k^{2} + 16 = 25\) and \(k^{2} = 9\).
  3. Touching internally: \(d = 3 - 2 = 1\), so \(k^{2} + 16 = 1\), \(k^{2} = -15\): impossible.
  4. So they can only touch externally.
Answer: \(k = 3\) or \(k = -3\)
Your turn
How do the circles \(x^{2} + y^{2} = 36\) and \((x - 1)^{2} + (y - 1)^{2} = 4\) sit?
Examiner trap

Check BOTH conditions. \(d < r_1 + r_2\) on its own does not mean two points: radii \(6\) and \(2\) with \(d = \sqrt{2}\) gives \(d < 8\), yet one circle is inside the other. And find the radius from the general form as \(\sqrt{g^{2} + f^{2} - c}\): \(g^{2} + f^{2} - c\) is \(r^{2}\), not \(r\).

Section 02

Subtract to find the common chord.

When two circles meet at \(P\) and \(Q\), both equations are true at those points, so their difference is true there too. Write each circle in the form \(x^{2} + y^{2} + \dots = 0\) and subtract: the \(x^{2}\) and \(y^{2}\) terms cancel and a straight line is left. That line passes through \(P\) and \(Q\), so it is the common chord. Substitute it back into either circle to find the points. In Fig. 1 the line of centres \(AB\) is perpendicular to the chord \(PQ\) and bisects it, which gives a quick check.

Worked example Chord and points of Fig. 1
The circles \(C_1: x^{2} + y^{2} - 6x - 6y - 7 = 0\) and \(C_2: x^{2} + y^{2} - 14x - 14y + 49 = 0\) meet at \(P\) and \(Q\). Find the equation of the common chord and the coordinates of \(P\) and \(Q\).
  1. \(C_1 - C_2\): \((-6x + 14x) + (-6y + 14y) + (-7 - 49) = 0\), so \(8x + 8y - 56 = 0\), giving \(x + y = 7\).
  2. Substitute \(y = 7 - x\) into \(C_1\): \(x^{2} + (7 - x)^{2} - 6x - 6(7 - x) - 7 = 0\).
  3. \(x^{2} + 49 - 14x + x^{2} - 6x - 42 + 6x - 7 = 0\), so \(2x^{2} - 14x = 0\) and \(2x(x - 7) = 0\).
  4. \(x = 0\) gives \(y = 7\); \(x = 7\) gives \(y = 0\).
  5. Check: the centres \((3, 3)\) and \((7, 7)\) give gradient \(1\), the chord has gradient \(-1\), and \(1 \times (-1) = -1\).
Answer: \(x + y = 7\); \(P(0, 7)\) and \(Q(7, 0)\)
Worked example Exact length of a common chord
Find the exact length of the common chord of \(x^{2} + y^{2} = 16\) and \(x^{2} + y^{2} - 8x = 0\).
  1. Subtract: \((x^{2} + y^{2} - 16) - (x^{2} + y^{2} - 8x) = 0\), so \(8x - 16 = 0\) and the chord is \(x = 2\).
  2. Substitute into the first circle: \(4 + y^{2} = 16\), so \(y^{2} = 12\) and \(y = \pm 2\sqrt{3}\).
  3. The ends are \((2, 2\sqrt{3})\) and \((2, -2\sqrt{3})\).
  4. Length \(= 2\sqrt{3} - (-2\sqrt{3}) = 4\sqrt{3}\).
Answer: \(4\sqrt{3}\)
Your turn
Find the points where \(x^{2} + y^{2} = 10\) meets \(x^{2} + y^{2} - 4x - 2y = 0\).
Examiner trap

Divide first if a circle is written as \(2x^{2} + 2y^{2} + \dots = 0\); otherwise the \(x^{2}\) and \(y^{2}\) terms do not cancel. Subtract EVERY term, including the constants: \(-7 - (+49) = -56\), not \(-7 + 49\). And the subtraction always produces a line, even for circles that never meet, so decide first (or check that the quadratic has real roots) that there is a chord at all.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Two circles have radii 3 and 5 and their centres are 8 units apart. How do they sit?

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Two circles have radii 2 and 7 and their centres are 4 units apart. How do they sit?

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Find the common chord of \(x^{2} + y^{2} - 9 = 0\) and \(x^{2} + y^{2} - 2x - 7 = 0\).

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Circles with centres \((0, 0)\) and \((6, 0)\) and radii 2 and 4 touch. Find the point of contact.

Section 03

Touching circles and the point of contact.

When two circles touch, the point of contact lies on the line of centres. If they touch externally, it divides \(AB\) in the ratio \(r_1 : r_2\), so it is the fraction \(\dfrac{r_1}{r_1 + r_2}\) of the way from \(A\) to \(B\). If they touch internally, it is on the line \(AB\) extended beyond the centre of the smaller circle, at distance \(r_1\) from the centre \(A\) of the larger one. Subtracting the equations of two touching circles gives their common tangent at the point of contact. A circle that touches both positive axes has centre \((r, r)\) and radius \(r\).

Worked example Point of contact
Show that the circles \(x^{2} + y^{2} - 2x - 4y - 11 = 0\) and \(x^{2} + y^{2} - 14x - 20y + 113 = 0\) touch, and find the point of contact.
  1. Centres \(A(1, 2)\) and \(B(7, 10)\); radii \(\sqrt{1 + 4 + 11} = 4\) and \(\sqrt{49 + 100 - 113} = 6\).
  2. \(d = \sqrt{6^{2} + 8^{2}} = 10 = 4 + 6\), so they touch externally.
  3. The contact point is \(\dfrac{4}{10}\) of the way from \(A\) to \(B\): \(\left(1 + \tfrac{2}{5}(6),\ 2 + \tfrac{2}{5}(8)\right) = \left(\tfrac{17}{5}, \tfrac{26}{5}\right)\).
  4. Check: subtracting the equations gives \(12x + 16y - 124 = 0\), i.e. \(3x + 4y = 31\), and \(3\left(\tfrac{17}{5}\right) + 4\left(\tfrac{26}{5}\right) = \tfrac{51 + 104}{5} = 31\).
Answer: They touch externally at \(\left(\frac{17}{5}, \frac{26}{5}\right)\)
Worked example A circle touching both axes and a circle
A circle lies in the first quadrant, touches both axes and touches the circle \(x^{2} + y^{2} = 4\) from the inside. Find its exact radius.
  1. Touching both axes: centre \((r, r)\), radius \(r\), so its distance from \(O\) is \(\sqrt{r^{2} + r^{2}} = r\sqrt{2}\).
  2. Touching internally (radii \(2\) and \(r\)): \(d = 2 - r\), so \(r\sqrt{2} = 2 - r\).
  3. \(r(\sqrt{2} + 1) = 2\), so \(r = \dfrac{2}{\sqrt{2} + 1}\).
  4. Rationalise: \(r = \dfrac{2(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \dfrac{2(\sqrt{2} - 1)}{1}\).
Answer: \(r = 2\sqrt{2} - 2\)
Your turn
Find the point of contact of \(x^{2} + y^{2} = 1\) and \((x - 3)^{2} + (y - 4)^{2} = 16\).
Examiner trap

Keep the ratio the right way round: the contact point is CLOSER to the centre of the smaller circle, at distance \(r_1\) from \(A\). Going \(\frac{r_1}{r_2}\) of the way (\(\frac{4}{6}\) above) instead of \(\frac{r_1}{r_1 + r_2}\) (\(\frac{4}{10}\)) is a common slip, and the midpoint of \(AB\) is the contact point only when the radii are equal.

Stage 2 · Exam

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Find all the values of \(k\) for which the circles \(x^{2} + y^{2} = 4\) and \((x - k)^{2} + y^{2} = 1\) touch.

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Find the equation of the common chord of \(x^{2} + y^{2} - 6x - 4y + 4 = 0\) and \(x^{2} + y^{2} - 2x - 8y + 8 = 0\).

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Find the exact length of the common chord of \(x^{2} + y^{2} = 12\) and \(x^{2} + y^{2} - 6x = 0\).

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The circles \(x^{2} + y^{2} = 9\) and \(x^{2} + y^{2} - 16x - 12y + 51 = 0\) touch externally. Find the point of contact.

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Centres measured, chords found.

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Six original Cambridge-style questions
Classifying pairs of circles, an unknown radius for touching, a common chord and its end points, a point of contact with the common tangent, and a circle touching both axes and another circle.