Before starting this question, make sure you are familiar with finding equations of straight lines (y = mx + c), and equations of perpendicular lines
It is a good idea to start with a sketch for a question like this
(Note that the radius of the circle, which is 2, is not needed to answer this question)

edexcel-tq-mis-2-15-h-5

First work out the gradient of the radius, OP, using gradient = $\frac{rise}{run} = \frac{y_{1} - y_{2}}{x_{1} - x_{2}}$
You can use your calculator, but be careful to input the square roots and fractions in the correct places
(If your calculator gives you the answer in exact form, it's best to work with this rather than the rounded decimal)

$gradient of O P = \frac{\frac{\sqrt{7}}{2}}{\frac{3}{2}} = \frac{\sqrt{7}}{2} \div \frac{3}{2}$

[1]

$= \frac{\sqrt{7}}{3}$

The tangent is perpendicular to the radius. Therefore the gradient of the tangent at P is the negative reciprocal of the gradient of OP

$m = - \frac{3}{\sqrt{7}} (or - \frac{3 \sqrt{7}}{7})$

[1]

The tangent takes the form $y = - \frac{3}{\sqrt{7}} x + c$ . Substitute the point P $(\frac{3}{2}, \frac{\sqrt{7}}{2})$ into $y = - \frac{3}{\sqrt{7}} x + c$

$\frac{\sqrt{7}}{2} = - \frac{3}{\sqrt{7}} \times \frac{3}{2} + c$

Solve to find $c$ . Remember you can use your calculator to do the calculations

$\frac{\sqrt{7}}{2} = - \frac{9}{2 \sqrt{7}} + c c = \frac{\sqrt{7}}{2} + \frac{9}{2 \sqrt{7}} = \frac{8}{\sqrt{7}} (o r = \frac{8 \sqrt{7}}{7})$

Write the equation of the tangent using your found values of $m$ and $c$

$y = - \frac{3}{\sqrt{7}} x + \frac{8}{\sqrt{7}} or y = - \frac{3 \sqrt{7}}{7} x + \frac{8 \sqrt{7}}{7}$ [1]
Other forms of the answer such as $y - \frac{\sqrt{7}}{2} = - \frac{3}{\sqrt{7}} (x - \frac{3}{2})$ or $y = - 1.13 . . . x + 3.02 . . .$ are permitted

Equation of a Circle (AQA GCSE Maths)

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