### Equation of circle

Question Sample Titled 'Equation of circle'

The coordinates of the points ${P}$ and ${Q}$ are ${\left({6},{5}\right)}$ and ${\left({14},{1}\right)}$ respectively.

 (a) It is given that ${L}$ is a line parallel to ${P}{Q}$ and passes through ${\left(-{6},{1}\right)}$.  (i) Find the equation of ${L}$ . (ii) Suppose that ${G}$ is a point lying on ${L}$ . Denote the ${y}$-coordinate of ${G}$ by ${k}$ . Let ${C}$ be the circle which is centred at ${G}$ and passes through ${P}$ . Prove that the equation of ${C}$ is ${x}^{{2}}+{y}^{{2}}+{4}{\left({k}+{2}\right)}{x}-{2}{k}{y}-{14}{k}-{109}={0}.$  (6 marks) (b) The coordinates of the point ${R}$ are ${\left(-{2},{1}\right)}$ . Using (a)(ii) , or otherwise, find the diameter of the circle centred at ${G}$ which passes through ${P}$ and ${R}$ .  (3 marks)

 (ai) Slope of ${P}{Q}$ $=\dfrac{{{5}-{1}}}{{{6}-{14}}}$ 1M  $=-\dfrac{{1}}{{2}}$ Equation of ${L}$: $\dfrac{{{y}-{1}}}{{{x}-{\left(-{6}\right)}}}$ $=-\dfrac{{1}}{{2}}$ 1M ${x}+{2}{y}+{4}$ $={0}$ 1A  (ii) ∵  ${G}$ lies on ${L}$ , ∴   The ${x}$-coordinate of ${G}$ is $-{2}{k}-{4}.$ 1A Equation of ${C}:$ ${\left[{x}-{\left(-{2}{k}-{4}\right)}\right]}^{{2}}+{\left({y}-{k}\right)}^{{2}}={\left(-{2}{k}-{4}-{6}\right)}^{{2}}+{\left({k}-{5}\right)}^{{2}}$ 1M Simplify, we have ${x}^{{2}}+{y}^{{2}}+{4}{\left({k}+{2}\right)}{x}-{2}{k}{y}-{14}{k}-{109}={0}$ 1A

 (b) Sub. ${\left(-{2},{1}\right)}$ into the equation of ${C}$ , ${\left(-{2}\right)}^{{2}}+{\left({1}\right)}^{{2}}+{4}{\left({k}+{2}\right)}{\left(-{2}\right)}-{2}{k}{\left({1}\right)}-{14}{k}-{109}$ $={0}$ $-{24}{k}$ $={120}$ ${k}$ $=-{5}$ 1A ∴   The centre of the circle is ${\left({6},-{5}\right)}.$ 1A   ∵  ${P}$ and ${G}$ are lying on the same vertical line. ∴   The radius of the circle$={5}-{\left(-{5}\right)}={10}$ ∴   The diameter$={20}$ units.  1A

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