### Equations of circle and straight line, mid-point of intersections

Question Sample Titled 'Equations of circle and straight line, mid-point of intersections'

The coordinates of the centre of the circle ${X}$ are ${\left({9},{3}\right)}$ . It is given that the ${x}$-axis is a tangent to ${X}$ .

 (a) Find the equation of ${X}$ . (2 marks) (b) The slope and the ${y}$-intercept of the straight line ${L}$ is $-{4}$ and ${h}$ respectively. If ${L}$ cuts ${X}$ at ${T}$ and ${G}$ , express the coordinates of the mid-point of ${T}{G}$ in terms of ${h}$ .  (5 marks)

 (a) Note that the radius of ${X}$ is ${3}$ . 1M can be absorbed Thus, the equation of ${X}$ is ${\left({x}-{9}\right)}^{{2}}+{\left({y}-{3}\right)}^{{2}}={3}^{{2}}$ . 1A or equivalent (b) The equation of ${L}$ is ${y}=-{4}{x}+{h}$ . 1M Putting the equation of ${L}$ into the equation of ${X}$, which can be rewritten as ${x}^{{2}}+{y}^{{2}}-{18}{x}-{6}{y}+{81}={0}$ , we have ${x}^{{2}}+{\left(-{4}{x}+{h}\right)}^{{2}}-{18}{x}-{6}{\left(-{4}{x}+{h}\right)}+{81}$ $={0}$ . 1M So, we have ${17}{x}^{{2}}+{\left({6}-{8}{h}\right)}{x}+{\left({h}^{{2}}-{6}{h}+{81}\right)}$ $={0}$ . The ${x}$-coordinate of the mid-point of ${T}{G}$ $=\dfrac{{\dfrac{{-{\left({6}-{8}{h}\right)}}}{{17}}}}{{2}}$ sum of roots 1M $=\dfrac{{-{3}+{4}{h}}}{{17}}$ 1A The ${y}$-coordinate of the mid-point of ${T}{G}$ $=-{4}{\left(\dfrac{{-{3}+{4}{h}}}{{17}}\right)}+{h}$ $=\dfrac{{{12}+{h}}}{{17}}$ Thus, the required coordinates are ${\left(\dfrac{{-{3}+{4}{h}}}{{17}},\dfrac{{{12}+{h}}}{{17}}\right)}$ . 1A f.t.

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