### Perpendicular bisector of radius of circle

Question Sample Titled 'Perpendicular bisector of radius of circle'

The equation of the circle ${C}$ is ${x}^{{2}}+{y}^{{2}}-{2}{x}-{6}{y}-{42}={0}$ . Denote the centre of the circle by ${G}$. ${F}{\left({7},{7}\right)}$ is a point lying on ${C}$.

 (a) Write down the coordinates of ${G}$. (1 mark) (b) The equation of the straight line ${L}$ is ${3}{x}+{2}{y}-{22}={0}$ . It is found that ${L}$ cuts ${C}$ at ${H}$ and ${K}$ . (i) Is ${L}$ the perpendicular bisector of ${F}{G}$ ? Explain your answer. (ii) Find the perimeter of the quadrilateral ${F}{H}{G}{K}$ in surd form.  (6 marks)

 (a) ${G}$ $={\left(-\dfrac{{-{2}}}{{2}},-\dfrac{{-{6}}}{{2}}\right)}$ 1A  $={\left({1},{3}\right)}$ (bi) Slope of ${L}$ $=-\dfrac{{3}}{{2}}$ Slope of ${F}{G}$ $=\dfrac{{{7}-{3}}}{{{7}-{1}}}$ 1M  $=\dfrac{{2}}{{3}}$ ∵  $\dfrac{{2}}{{3}}\times{\left(-\dfrac{{3}}{{2}}\right)}=-{1}$ ∴  ${F}{G}\bot{L}$ 1M The mid-point of ${F}{G}={\left(\dfrac{{{1}+{7}}}{{2}},\dfrac{{{3}+{7}}}{{2}}\right)}={\left({4},{5}\right)}$  Substitute ${\left({4},{5}\right)}$ into ${L}$, we have ${L}.{H}.{S}={3}{\left({4}\right)}+{2}{\left({5}\right)}-{22}={0}$ ∴  ${\left({4},{5}\right)}$ is a point on ${L}$ .  1M Thus, ${L}$ is the perpendicular bisector of ${F}{G}$ . 1A  (bii) ${F}{G}$ $=$ radius of ${C}$  $=\sqrt{{{\left({7}-{1}\right)}^{{2}}+{\left({7}-{3}\right)}^{{2}}}}$ 1M  $=\sqrt{{{52}}}$  $={2}\sqrt{{13}}$ ∴   Perimeter of ${F}{H}{G}{K}$ $={4}\times{2}\sqrt{{13}}$  $={8}\sqrt{{13}}$ $\text{units}$ 1A

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