2022 H2 Maths Paper 1 Q5 — Worked Solution

Part (a)

4 marks

Approach

a line is tangent to a circle exactly when the perpendicular distance from the centre to the line equals the radius. Writing $y = mx$ as $mx - y = 0$ lets us use the standard point-to-line distance formula directly.

The circle $(x + 8)^{2} + (y - 14)^{2} = 52$ has centre $( - 8,14)$ and radius $\sqrt{52}$ . The line $y = mx$ can be written $mx - y = 0$ . Since it is tangent to the circle, the perpendicular distance from the centre to the line equals the radius.

\frac{\left| m( - 8) - 14 \right|}{\sqrt{m^{2} + 1}} = \sqrt{52}

\Rightarrow \frac{| - 8m - 14|}{\sqrt{m^{2} + 1}} = \sqrt{52}

Squaring both sides,

\frac{(8m + 14)^{2}}{m^{2} + 1} = 52

\Rightarrow (8m + 14)^{2} = 52\left( m^{2} + 1 \right)

\Rightarrow 64m^{2} + 224m + 196 = 52m^{2} + 52

\Rightarrow 12m^{2} + 224m + 144 = 0

Dividing through by $4$ ,

3m^{2} + 56m + 36 = 0\ \ \left( \text{shown} \right)

To check your answer: dividing

12m^{2} + 224m + 144 = 0

4

gives

3m^{2} + 56m + 36 = 0

directly, matching the target equation ✓

Part (b)

4 marks

Approach

part (a) gives the gradients $m$ of the two tangents from the origin. Each point of tangency is the foot of the perpendicular from the circle’s centre to its tangent line, so once $m$ is known for each tangent, the corresponding point can be found directly without solving the circle and line simultaneously.

Solving $3m^{2} + 56m + 36 = 0$ by the quadratic formula, discriminant $\text{=}56^{2} - 4(3)(36) = 3136 - 432 = 2704 = 52^{2}$ ,

m = \frac{- 56 \pm 52}{6}

\Rightarrow m = - \frac{2}{3}\ \ \text{or}\ \ m = - 18

To check your answer: product of roots

\text{=}\frac{36}{3} = 12

, and

\left( - \frac{2}{3} \right)( - 18) = 12

✓. Sum of roots

\text{=} - \frac{56}{3}

, and

- \frac{2}{3} - 18 = - \frac{2 + 54}{3} = - \frac{56}{3}

✓

The point of tangency for a tangent $y = mx$ is the foot of the perpendicular from the centre $( - 8,14)$ to the line. Parametrizing the line as $(t,mt)$ , this foot is at

t = \frac{x_{0} + my_{0}}{1 + m^{2}},\quad\text{point} = (t,\ mt)

where $\left( x_{0},y_{0} \right) = ( - 8,14)$ .

For $m = - \frac{2}{3}$ (point $A$ ),

t = \frac{- 8 + \left( - \frac{2}{3} \right)(14)}{1 + \frac{4}{9}}

= \frac{- 8 - \frac{28}{3}}{\frac{13}{9}}

= \frac{- \frac{52}{3}}{\frac{13}{9}}

= - \frac{52}{3} \times \frac{9}{13}

= - 12

so $y = - \frac{2}{3}( - 12) = 8$ , giving $A = ( - 12,\ 8)$ .

For $m = - 18$ (point $B$ ),

t = \frac{- 8 + ( - 18)(14)}{1 + 324}

= \frac{- 8 - 252}{325}

= - \frac{260}{325}

= - \frac{4}{5}

so $y = - 18\left( - \frac{4}{5} \right) = \frac{72}{5}$ , giving $B = \left( - \frac{4}{5},\ \frac{72}{5} \right)$ .

To check your answer: on the circle,

A = ( - 12,8)

gives

( - 12 + 8)^{2} + (8 - 14)^{2} = 16 + 36 = 52

✓, and on the line,

8 = - \frac{2}{3}( - 12) = 8

✓. For

B = \left( - \frac{4}{5},\frac{72}{5} \right)

\left( - \frac{4}{5} + 8 \right)^{2} + \left( \frac{72}{5} - 14 \right)^{2} = \left( \frac{36}{5} \right)^{2} + \left( \frac{2}{5} \right)^{2} = \frac{1296 + 4}{25} = \frac{1300}{25} = 52

✓, and on the line

- 18\left( - \frac{4}{5} \right) = \frac{72}{5}

✓

$\therefore$ $A = ( - 12,\ 8)$ and $B = \left( - \frac{4}{5},\ \frac{72}{5} \right)$ .

2022 H2 Maths Paper 1, Question 5

Part (a)

Part (b)

This is how we mark, and how we teach.