8 Coordinate geometry of the circle | IGCSE Additional Maths 0606

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🎯What you need to be able to do

Use \( (x - a)^{2} + (y - b)^{2} = r^{2} \), and find the centre and radius from \( x^{2} + y^{2} + 2gx + 2fy + c = 0 \).
Find where a line meets a circle, and decide whether it is a chord, a tangent or misses.
Find the equation of a tangent to a circle at a given point, without calculus.
Decide whether two circles intersect, touch or do not meet; find their points of intersection and the equation of the common chord.

📚The mathematics

Two forms of the equation

A circle with centre \( (a, b) \) and radius \(r\) is \( (x - a)^{2} + (y - b)^{2} = r^{2} \) — this one is in the List of formulas. Expanded, every circle looks like

\[ x^{2} + y^{2} + 2gx + 2fy + c = 0, \qquad \text{centre } (-g, -f), \quad r = \sqrt{g^{2} + f^{2} - c} \]

Rather than memorise that, complete the square in \(x\) and in \(y\). It always works and it shows the working the mark scheme wants. Note that an equation of a circle has equal coefficients of \( x^{2} \) and \( y^{2} \) and no \( xy \) term; if they differ, divide through first.

A line and a circle

Substitute the line into the circle to get a quadratic. Then, exactly as in Topic 2:

\( b^{2} - 4ac > 0 \): a chord, two points

\( b^{2} - 4ac = 0 \): a tangent, one point

\( b^{2} - 4ac < 0 \): the line does not meet the circle

Tangents

A tangent is perpendicular to the radius at the point of contact. So: find the gradient of the radius from the centre to the point, take the negative reciprocal, and use \( y - y_{1} = m(x - x_{1}) \). No calculus is expected. The perpendicular from the centre to any chord bisects the chord, which gives the second most useful right angle in circle questions.

The length of a tangent from an outside point \(P\) comes from Pythagoras: if \(C\) is the centre and \(T\) the point of contact, \( PT^{2} = PC^{2} - r^{2} \).

Two circles

Compare the distance \(d\) between the centres with the radii \( r_{1} \ge r_{2} \):

\( d > r_{1} + r_{2} \): separate

\( d = r_{1} + r_{2} \): touch externally

\( r_{1} - r_{2} < d < r_{1} + r_{2} \): intersect at two points

\( d = r_{1} - r_{2} \): touch internally

\( d < r_{1} - r_{2} \): one inside the other

To find where two circles meet, write both in expanded form and subtract: the \( x^{2} \) and \( y^{2} \) terms cancel, leaving a straight line. That line is the common chord (or the common tangent, if they touch). Then solve it simultaneously with either circle.

The circles x squared plus y squared = 25 and (x minus 6) squared plus y squared = 13 overlapping. Subtracting their equations gives the vertical line x = 4, which passes through both intersection points (4, 3) and (4, minus 3). — Subtracting \( x^{2} + y^{2} = 25 \) and \( x^{2} - 12x + 36 + y^{2} = 13 \) leaves \( 12x - 36 = 12 \), i.e. \( x = 4 \): the common chord.

✏️Worked example

The circle \( C_{1} \) has equation \( x^{2} + y^{2} - 6x + 4y - 12 = 0 \). (a) Find the centre and radius of \( C_{1} \). [3] (b) Show that \( P(6, 2) \) lies on \( C_{1} \), and find the equation of the tangent at \(P\). [4] (c) The circle \( C_{2} \) has centre \( (9, 6) \) and radius 5. Show that \( C_{1} \) and \( C_{2} \) touch at \(P\). [3]

(a) \( (x - 3)^{2} - 9 + (y + 2)^{2} - 4 - 12 = 0 \), so \( (x - 3)^{2} + (y + 2)^{2} = 25 \). Centre \( (3, -2) \), radius 5.

(b) \( (6 - 3)^{2} + (2 + 2)^{2} = 9 + 16 = 25 \), so \(P\) is on the circle. The radius to \(P\) has gradient \( \dfrac{2 - (-2)}{6 - 3} = \tfrac{4}{3} \), so the tangent has gradient \( -\tfrac{3}{4} \):

\[ y - 2 = -\tfrac{3}{4}(x - 6) \;\Longrightarrow\; 3x + 4y = 26 \]

(c) The distance between the centres is \( \sqrt{(9 - 3)^{2} + (6 + 2)^{2}} = \sqrt{100} = 10 = 5 + 5 \), so the circles touch externally. The point of contact is on the line of centres, 5 from each centre — the midpoint of \( (3, -2) \) and \( (9, 6) \), which is \( (6, 2) = P \).

Two circles of radius 5, centred at (3, minus 2) and (9, 6), touching at P(6, 2). The line of centres passes through P, and the common tangent 3x + 4y = 26 touches both circles at P, perpendicular to the line of centres. — The circles touch at \(P\), the midpoint of the line of centres; \( 3x + 4y = 26 \) is a tangent to both.

Check it. The tangent should be perpendicular to the line of centres, whose gradient is \( \tfrac{8}{6} = \tfrac{4}{3} \): and \( \tfrac{4}{3} \times \left(-\tfrac{3}{4}\right) = -1 \) ✓.

Sign of the centre. \( (y + 2)^{2} \) means \( b = -2 \). Giving the centre as \( (3, 2) \) loses the mark and everything that follows from it.

📝Practise

Circle questions usually appear on Paper 1; one here needs a calculator.

1. (No calculator.) Find the centre and radius of the circle \( x^{2} + y^{2} + 8x - 10y + 5 = 0 \). [3]

\( (x + 4)^{2} - 16 + (y - 5)^{2} - 25 + 5 = 0 \Rightarrow (x + 4)^{2} + (y - 5)^{2} = 36 \). Centre \( (-4, 5) \), radius 6.

2. (No calculator.) \( A(-1, 4) \) and \( B(5, -4) \) are the ends of a diameter of a circle. Find the equation of the circle. [3]

Centre = midpoint \( (2, 0) \). \( AB = \sqrt{36 + 64} = 10 \), so \( r = 5 \). Equation \( (x - 2)^{2} + y^{2} = 25 \).

3. (No calculator.) Show that the line \( 3x - 4y = 25 \) is a tangent to the circle \( x^{2} + y^{2} = 25 \), and find the point of contact. [4]

\( x = \dfrac{25 + 4y}{3} \). Substituting and multiplying by 9: \( 625 + 200y + 16y^{2} + 9y^{2} = 225 \Rightarrow 25y^{2} + 200y + 400 = 0 \Rightarrow (y + 4)^{2} = 0 \). A repeated root, so the line is a tangent; \( y = -4 \), \( x = 3 \): contact at \( (3, -4) \).

4. (No calculator.) Find the values of \(k\) for which the line \( y = x + k \) is a tangent to \( x^{2} + y^{2} = 8 \). [3]

\( x^{2} + (x + k)^{2} = 8 \Rightarrow 2x^{2} + 2kx + k^{2} - 8 = 0 \). Tangent: \( 4k^{2} - 8(k^{2} - 8) = 0 \Rightarrow 64 - 4k^{2} = 0 \Rightarrow k = \pm 4 \).

5. (No calculator.) Show that the circles \( (x - 1)^{2} + (y - 2)^{2} = 9 \) and \( (x - 7)^{2} + (y + 6)^{2} = 36 \) do not meet. [3]

Centres \( (1, 2) \) and \( (7, -6) \): \( d = \sqrt{36 + 64} = 10 \). Radii 3 and 6, so \( r_{1} + r_{2} = 9 < 10 \): the circles are separate and do not meet.

6. (Calculator.) The circles \( x^{2} + y^{2} - 4x - 6y - 12 = 0 \) and \( x^{2} + y^{2} - 12x - 6y + 20 = 0 \) intersect at \(A\) and \(B\). Find the equation of the common chord and the length \(AB\). [5]

Subtract: \( 8x - 32 = 0 \), so the common chord is \( x = 4 \). In the first circle: \( 16 + y^{2} - 16 - 6y - 12 = 0 \Rightarrow y^{2} - 6y - 12 = 0 \Rightarrow y = 3 \pm \sqrt{21} \). So \( AB = 2\sqrt{21} = 9.17 \).

7. (No calculator.) Find the length of the tangent from \( P(1, 7) \) to the circle \( (x - 4)^{2} + (y - 3)^{2} = 9 \). [3]

Centre \( C(4, 3) \), \( r = 3 \). \( PC^{2} = 9 + 16 = 25 \). The tangent meets the radius at a right angle, so \( PT^{2} = 25 - 9 = 16 \) and \( PT = 4 \).

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

GeoGebra — drag the centre of one circle and watch the common chord appear, shrink and become a tangent
Desmos — enter a circle in expanded form to check your completed square