IGCSE · Practice questions · A circle and a straight line

Cut, touch or miss.

Six original Cambridge-style questions on a circle and a straight line: substituting to find points of intersection, the discriminant test for a chord, a tangent or no intersection, constants for tangency, the perpendicular from the centre and chord lengths, 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

Substitute, collect, then read the discriminant.

Rearrange the line for \(y\) or \(x\), choosing the way that avoids fractions.
Substitute into the circle with brackets and collect to \(ax^{2} + bx + c = 0\).
Decide: \(b^{2} - 4ac > 0\) chord, \(= 0\) tangent, \(< 0\) misses.
Pair up: find each other coordinate from the line.
Perpendicular from C: its foot is the midpoint of a chord; compare \(CF^{2}\) with \(r^{2}\), and half the chord is \(\sqrt{r^{2} - CM^{2}}\).

Recall

[3 marks]

Determine whether the line \(y = x + 3\) is a chord of, a tangent to, or does not meet the circle \(x^{2} + y^{2} = 4\).

Substitute: \(x^{2} + (x + 3)^{2} = 4\), so \(2x^{2} + 6x + 9 - 4 = 0\), that is \(2x^{2} + 6x + 5 = 0\). ✓
Discriminant: \(6^{2} - 4(2)(5) = 36 - 40 = -4\). ✓
\(-4 < 0\), so there are no real roots: the line does not meet the circle. ✓

Calculation

[6 marks]

The line \(7x + y = 34\) meets the circle \(x^{2} + y^{2} - 2x - 4y - 20 = 0\) at A and B.

(a) Find the coordinates of A and B.

(b) Find the coordinates of the midpoint of AB and the exact length of AB.

(a) \(y = 34 - 7x\). Substitute: \(x^{2} + (34 - 7x)^{2} - 2x - 4(34 - 7x) - 20 = 0\). ✓
\(x^{2} + 1156 - 476x + 49x^{2} - 2x - 136 + 28x - 20 = 0\), so \(50x^{2} - 450x + 1000 = 0\). ✓
Divide by 50: \(x^{2} - 9x + 20 = 0\), so \((x - 4)(x - 5) = 0\). ✓
\(x = 4\): \(y = 34 - 28 = 6\); \(x = 5\): \(y = 34 - 35 = -1\). So A\((4,\ 6)\) and B\((5,\ -1)\). ✓
(b) Midpoint \(\left(\dfrac{4 + 5}{2},\ \dfrac{6 + (-1)}{2}\right) = (4.5,\ 2.5)\). ✓
\(AB = \sqrt{(5 - 4)^{2} + (-1 - 6)^{2}} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}\). ✓

Calculation

[4 marks]

Show that the line \(4x - 3y = 30\) is a tangent to the circle \(x^{2} + y^{2} = 36\), and find the coordinates of the point of contact.

\(y = \dfrac{4x - 30}{3}\). Substitute and multiply by 9: \(9x^{2} + (4x - 30)^{2} = 324\). ✓
\(9x^{2} + 16x^{2} - 240x + 900 - 324 = 0\), so \(25x^{2} - 240x + 576 = 0\). ✓
Discriminant: \(240^{2} - 4(25)(576) = 57600 - 57600 = 0\), so the line is a tangent. ✓
\(25x^{2} - 240x + 576 = (5x - 24)^{2} = 0\), so \(x = \dfrac{24}{5}\). ✓
\(y = \dfrac{4 \times \frac{24}{5} - 30}{3} = \dfrac{\frac{96 - 150}{5}}{3} = -\dfrac{18}{5}\). The point of contact is \(\left(\dfrac{24}{5},\ -\dfrac{18}{5}\right)\). ✓

Analysis

[6 marks]

The circle C has equation \((x - 3)^{2} + (y - 1)^{2} = 8\).

(a) Find the values of \(k\) for which the line \(y = x + k\) is a tangent to C, and find the point of contact in each case.

(b) Write down the set of values of \(k\) for which the line \(y = x + k\) is a chord of C.

(a) Substitute: \((x - 3)^{2} + (x + k - 1)^{2} = 8\). ✓
\(x^{2} - 6x + 9 + x^{2} + 2(k - 1)x + (k - 1)^{2} - 8 = 0\), so \(2x^{2} + (2k - 8)x + (k^{2} - 2k + 2) = 0\). ✓
Tangent: \((2k - 8)^{2} - 8(k^{2} - 2k + 2) = 0\), so \(4k^{2} - 32k + 64 - 8k^{2} + 16k - 16 = 0\), giving \(-4k^{2} - 16k + 48 = 0\). ✓
\(k^{2} + 4k - 12 = 0\), so \((k + 6)(k - 2) = 0\) and \(k = 2\) or \(k = -6\). ✓
\(k = 2\): \(2x^{2} - 4x + 2 = 2(x - 1)^{2} = 0\), so the contact point is \((1,\ 3)\). ✓
\(k = -6\): \(2x^{2} - 20x + 50 = 2(x - 5)^{2} = 0\), so the contact point is \((5,\ -1)\). ✓
(b) A chord needs \(-4k^{2} - 16k + 48 > 0\), that is \(k^{2} + 4k - 12 < 0\), so \(-6 < k < 2\) (between the two tangents). ✓

Analysis

[6 marks]

The circle \(x^{2} + y^{2} - 8x - 6y = 0\) has centre C. The line L has equation \(y = 2x - 15\).

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

(b) Find the coordinates of F, the foot of the perpendicular from C to L.

(a) \((x - 4)^{2} + (y - 3)^{2} = 16 + 9 = 25\): centre C\((4,\ 3)\), radius 5. ✓
(b) L has gradient 2, so the perpendicular through C has gradient \(-\tfrac{1}{2}\): \(y - 3 = -\tfrac{1}{2}(x - 4)\), that is \(y = 5 - \tfrac{1}{2}x\). ✓
Solve with L: \(2x - 15 = 5 - \tfrac{1}{2}x\), so \(\tfrac{5}{2}x = 20\), \(x = 8\) and \(y = 1\). F is \((8,\ 1)\). ✓
(c) \(CF^{2} = (8 - 4)^{2} + (1 - 3)^{2} = 16 + 4 = 20 < 25\), so \(CF < r\) and L is a chord. ✓
F is the midpoint of the chord, so \(\left(\text{half chord}\right)^{2} = 25 - 20 = 5\). ✓
Chord length \(= 2\sqrt{5}\). (Check: L meets the circle at \((7,\ -1)\) and \((9,\ 3)\), which are \(\sqrt{4 + 16} = 2\sqrt{5}\) apart.) ✓

Analysis

[6 marks]

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

(a) Show that \(m^{2} = 4\).

(b) For each value of \(m\), find the point of contact.

(a) Substitute: \(x^{2} + (mx + 5)^{2} = 5\), so \((1 + m^{2})x^{2} + 10mx + 20 = 0\). ✓
Tangent: \((10m)^{2} - 4(1 + m^{2})(20) = 0\), so \(100m^{2} - 80 - 80m^{2} = 0\), \(20m^{2} = 80\) and \(m^{2} = 4\). ✓
(b) \(m = 2\): \(5x^{2} + 20x + 20 = 5(x + 2)^{2} = 0\), so \(x = -2\), \(y = 1\): the point \((-2,\ 1)\). ✓
\(m = -2\): \(5x^{2} - 20x + 20 = 5(x - 2)^{2} = 0\), so \(x = 2\), \(y = 1\): the point \((2,\ 1)\). ✓
(c) Both contact points have \(y = 1\), so the base is the distance from \((-2,\ 1)\) to \((2,\ 1)\), which is 4, and the height from P\((0,\ 5)\) is \(5 - 1 = 4\). ✓
Area \(= \tfrac{1}{2} \times 4 \times 4 = 8\) 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.