IGCSE · Practice questions · Tangents to a circle

Perpendicular, then touch.

Six original Cambridge-style questions on tangents to a circle: tangents and normals at a known point, showing a line is a tangent with the discriminant, finding a constant, tangent lengths by Pythagoras and the tangents from an external point, all 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

Radius, perpendicular, point of contact.

Centre first: complete the square to read \(C\) and \(r^{2}\).
Known point: gradient of \(CT\), negative reciprocal, line through \(T\). The normal is the line \(CT\).
Is it a tangent? Substitute the line, discriminant \(= 0\), repeated root gives the point of contact.
Tangent length: \(PT^{2} = PC^{2} - r^{2}\).
From an external point: line with gradient \(m\) through \(P\), discriminant \(= 0\), solve the quadratic in \(m\).

Recall

[3 marks]

The point \(A(1, 6)\) lies on the circle \((x + 1)^{2} + (y - 3)^{2} = 13\). Find the equation of the tangent to the circle at \(A\), giving your answer in the form \(ax + by = c\).

Centre \(C(-1, 3)\). Gradient of \(CA = \dfrac{6 - 3}{1 - (-1)} = \dfrac{3}{2}\). ✓
Tangent gradient \(-\dfrac{2}{3}\): \(y - 6 = -\dfrac{2}{3}(x - 1) \Rightarrow 3y - 18 = -2x + 2\). ✓
\(2x + 3y = 20\). ✓

Calculation

[5 marks]

The circle \(x^{2} + y^{2} + 2x - 8y - 8 = 0\) passes through the point \(B(3, 7)\).

(a) Find the centre and radius of the circle.

(b) Find the equation of the normal at \(B\) and show that it passes through the centre.

(a) \((x + 1)^{2} + (y - 4)^{2} = 1 + 16 + 8 = 25\): centre \((-1, 4)\), radius 5. ✓
(b) Gradient of \(CB = \dfrac{7 - 4}{3 + 1} = \dfrac{3}{4}\). Normal: \(y - 7 = \dfrac{3}{4}(x - 3) \Rightarrow 3x - 4y + 19 = 0\). ✓
At \((-1, 4)\): \(-3 - 16 + 19 = 0\), so the normal passes through the centre. ✓
(c) Tangent gradient \(-\dfrac{4}{3}\): \(y - 7 = -\dfrac{4}{3}(x - 3) \Rightarrow 4x + 3y = 33\). ✓
At \(y = 0\): \(x = \dfrac{33}{4}\), the point \(\left(\dfrac{33}{4}, 0\right)\). ✓

Proof

[4 marks]

Show that the line \(y = x - 7\) is a tangent to the circle \(x^{2} + y^{2} - 4x + 2y - 3 = 0\), and find the coordinates of the point of contact.

Substitute: \(x^{2} + (x - 7)^{2} - 4x + 2(x - 7) - 3 = 0\). ✓
\(2x^{2} - 16x + 32 = 0 \Rightarrow x^{2} - 8x + 16 = 0\). ✓
Discriminant \((-8)^{2} - 4(1)(16) = 0\): a repeated root, so the line is a tangent. ✓
\((x - 4)^{2} = 0\) gives \(x = 4\), \(y = 4 - 7 = -3\): the point \((4, -3)\). ✓

Analysis

[5 marks]

The line \(y = kx\) is a tangent to the circle \((x - 5)^{2} + y^{2} = 5\).

(a) Find the possible values of \(k\).

(b) Find the point of contact for the positive value of \(k\).

(a) Substitute: \(x^{2} - 10x + 25 + k^{2}x^{2} = 5 \Rightarrow (1 + k^{2})x^{2} - 10x + 20 = 0\). ✓
Discriminant \(= 0\): \(100 - 80(1 + k^{2}) = 0 \Rightarrow k^{2} = \dfrac{1}{4}\), so \(k = \pm \dfrac{1}{2}\). ✓
(b) \(k = \dfrac{1}{2}\): \(\dfrac{5}{4}x^{2} - 10x + 20 = 0 \Rightarrow x^{2} - 8x + 16 = 0\), so \(x = 4\), \(y = 2\): the point \((4, 2)\). ✓
(c) \(OC^{2} = 25\), \(r^{2} = 5\), so \(OT^{2} = 20\) and \(OT = 2\sqrt{5}\) (check: \(4^{2} + 2^{2} = 20\)). ✓

Analysis

[6 marks]

The circle \(x^{2} + y^{2} - 2x - 4y = 0\) and the point \(P(4, 1)\) are given.

(a) Show that \(P\) lies outside the circle.

(b) A line through \(P\) has gradient \(m\). Show that the line is a tangent to the circle when \(2m^{2} + 3m - 2 = 0\).

(a) \((x - 1)^{2} + (y - 2)^{2} = 5\): centre \((1, 2)\), \(r^{2} = 5\). \(PC^{2} = 9 + 1 = 10 > 5\), so \(P\) is outside. ✓
(b) The line is \(y = mx + 1 - 4m\). Substituting gives \((1 + m^{2})x^{2} - 2(4m^{2} + m + 1)x + 16m^{2} + 8m - 3 = 0\). ✓
Discriminant: \(4(4m^{2} + m + 1)^{2} - 4(1 + m^{2})(16m^{2} + 8m - 3) = 4(-4m^{2} - 6m + 4) = -8(2m^{2} + 3m - 2)\). ✓
Tangent when this is \(0\), that is when \(2m^{2} + 3m - 2 = 0\). ✓
(c) \((2m - 1)(m + 2) = 0\): \(m = \dfrac{1}{2}\) gives \(x - 2y = 2\); \(m = -2\) gives \(y = -2x + 9\). ✓
The gradients multiply to \(\dfrac{1}{2} \times (-2) = -1\), so the tangents are perpendicular. ✓

Analysis

[6 marks]

The points \(A(4, 2)\) and \(B(5, -5)\) lie on the circle \((x - 1)^{2} + (y + 2)^{2} = 25\), with centre \(C\). The tangents at \(A\) and \(B\) meet at \(P\).

(a) Find the equations of the two tangents.

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

(a) \(C(1, -2)\). Gradient of \(CA = \dfrac{4}{3}\), so the tangent at \(A\) is \(y - 2 = -\dfrac{3}{4}(x - 4)\), that is \(3x + 4y = 20\). ✓
Gradient of \(CB = -\dfrac{3}{4}\), so the tangent at \(B\) is \(y + 5 = \dfrac{4}{3}(x - 5)\), that is \(4x - 3y = 35\). ✓
(b) \(9x + 12y = 60\) and \(16x - 12y = 140\) add to \(25x = 200\), so \(x = 8\) and \(y = -1\): \(P(8, -1)\). ✓
(c) \(PA^{2} = 16 + 9 = 25\) and \(PB^{2} = 9 + 16 = 25\), so \(PA = PB = 5\). ✓
\(CA = CB = 5\) and both angles at \(A\) and \(B\) are right angles; \(CA\) and \(CB\) have gradients \(\dfrac{4}{3}\) and \(-\dfrac{3}{4}\), so the angle \(ACB\) is also a right angle. ✓
\(CAPB\) is a square of side 5, so its area is 25 square units. ✓

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.