It helps to visualise the problem if you sketch the information given

cie-spec1-q10

First write the equation of the circle using the information given. ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ where $(h, k)$ is the centre and $r$ is the radius

${(x - 2)}^{2} + {(y + 4)}^{2} = 3^{2}$

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Now substitute $y = 2 x - 3$ into the equation of the circle to find the two possible $x$ values of the intersection of the line and circle

$\begin{array}{rcl} {(x - 2)}^{2} + {(2 x - 3 + 4)}^{2} & = & 3^{2} \end{array}$

[1]

Solve for $x$

$\begin{array}{rcl} x^{2} - 4 x + 4 + {(2 x + 1)}^{2} & = & 9 \\ x^{2} - 4 x + 4 + 4 x^{2} + 4 x + 1 & = & 9 \\ 5 x^{2} + 5 & = & 9 \\ 5 x^{2} & = & 4 \\ x^{2} & = & \frac{4}{5} \\ x & = & \pm \frac{2}{\sqrt{5}} \end{array}$

Substitute both possible values into the equation of the straight line in order to find pairs of coordinate points for A and B

$\begin{array}{rcl} y & = & 2 (\frac{2}{\sqrt{5}}) - 3 \\ = & \frac{4}{\sqrt{5}} - 3 \\ A & = & (\frac{2}{\sqrt{5}}, \frac{4}{\sqrt{5}} - 3) \end{array}$ $\begin{array}{rcl} y & = & 2 (- \frac{2}{\sqrt{5}}) - 3 \\ = & \frac{- 4}{\sqrt{5}} - 3 \\ B & = & (\frac{- 2}{\sqrt{5}}, \frac{- 4}{\sqrt{5}} - 3) \end{array}$

one mark for each correct pair of coordinates [2]

(Note that A and B here correspond to the diagram we drew at the beginning of the question but are of course interchangeable depending on your diagram)

Find AB using Pythagoras' theorem

$\begin{array}{rcl} A B & = & \sqrt{{(\frac{2}{\sqrt{5}} - \frac{- 2}{\sqrt{5}})}^{2} + {(\frac{4}{\sqrt{5}} - 3 - (\frac{- 4}{\sqrt{5}} - 3))}^{2}} \end{array}$

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$\begin{array}{rcl} = & \sqrt{{(\frac{4}{\sqrt{5}})}^{2} + {(\frac{8}{\sqrt{5}})}^{2}} \\ = & \sqrt{\frac{16}{5} + \frac{64}{5}} \\ = & \sqrt{16} \\ = & 4 \end{array}$

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XY is a perpendicular bisector of AB which is a chord. Therefore XY passes through the centre and therefore it is the diameter of the circle, so no real working is required! The radius is 3 so

$\begin{array}{rcl} X Y & = & 6 \end{array}$

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The area of the kite is

Area $\begin{array}{rcl} = & \frac{1}{2} \times 4 \times 6 \end{array}$

12 [1]

Coordinate Geometry of the Circle (CIE IGCSE Additional Maths)

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