### Find the equation of a circle given two points by considering the perpendicular bisector of two points.

Question Sample Titled 'Find the equation of a circle given two points by considering the perpendicular bisector of two points.'

The coodinates of the points ${P}$ and ${Q}$ are ${\left(-{2},{2}\right)}$ and ${\left({7},-{1}\right)}$ respectively.

 (a) Let ${L}$ be the perpendicular bisector of ${P}{Q}$ . (i) Find the equation of ${L}$ . (ii) Suppose that ${G}$ is a point lying on ${L}$. Denote the ${x}$-coordinate of ${G}$ be ${h}$ . Let ${C}$ be the circle which is centred at ${G}$ and passes through ${P}$ and ${Q}$ . Prove that the equation of ${C}$ is ${x}^{{2}}+{y}^{{2}}-{2}{h}{x}+{\left(-{6}{h}+{14}\right)}{y}+{\left({8}{h}-{36}\right)}={0}$ .  (6 marks) (b) The coordinates of point ${R}$ are ${\left(-{25},-{1}\right)}$ . Using (a)(ii), or otherwise, find the diameter of the circle which passes through ${P}$ , ${Q}$ and ${R}$ . (3 marks)

 (a)(i) The mid-point of ${P}{Q}$ $={\left(\dfrac{{5}}{{2}},\dfrac{{1}}{{2}}\right)}$ The slope of ${P}{Q}$ $=\dfrac{{-{1}-{2}}}{{{7}+{2}}}$ 1M $=-\dfrac{{1}}{{3}}$ The slope of ${L}$ $=-{1}\div{\left(-\dfrac{{1}}{{3}}\right)}$ $={3}$ The equation of ${L}$ is  ${y}-\dfrac{{1}}{{2}}={3}{\left({x}-\dfrac{{5}}{{2}}\right)}$ 1M ${y}={3}{x}-{7}$ 1A or equivalent  (a)(ii) Let ${k}$ be the ${y}$-coordinate of ${G}$ . By (a)(i), we have ${k}={3}{h}-{7}$ . 1M So, the coordinates of ${G}$ is ${\left({h},{3}{h}-{7}\right)}$ . The equation of ${C}$ is  ${\left({x}-{h}\right)}^{{2}}+{\left[{y}-{\left({3}{h}-{7}\right)}\right]}^{{2}}$ $={\left(-{2}-{h}\right)}^{{2}}+{\left[{2}-{\left({3}{h}-{7}\right)}\right]}^{{2}}$ 1M ${\left({x}-{h}\right)}^{{2}}-{\left(-{2}-{h}\right)}^{{2}}$ $={\left[{2}-{\left({3}{h}-{7}\right)}\right]}^{{2}}-{\left[{y}-{\left({3}{h}-{7}\right)}\right]}^{{2}}$ ${\left({x}-{2}{h}-{2}\right)}{\left({x}+{2}\right)}$ $={\left[{2}+{y}-{2}{\left({3}{h}-{7}\right)}\right]}{\left(-{y}+{2}\right)}$ ${a}^{{2}}+{b}^{{2}}={\left({a}+{b}\right)}{\left({a}-{b}\right)}$ ${x}^{{2}}-{2}{h}{x}-{2}{x}+{2}{x}-{4}{h}-{4}$ $=-{2}{y}-{y}^{{2}}+{2}{\left({3}{h}-{7}\right)}{y}+{4}+{2}{y}-{4}{\left({3}{h}-{7}\right)}$ ${x}^{{2}}-{2}{h}{x}-{4}{h}-{4}$ $=-{y}^{{2}}+{2}{\left({3}{h}-{7}\right)}{y}+{4}-{4}{\left({3}{h}-{7}\right)}$ ${x}^{{2}}+{y}^{{2}}-{2}{h}{x}-{2}{\left({3}{h}-{7}\right)}{y}-{4}{h}-{4}-{4}+{4}{\left({3}{h}-{7}\right)}={0}$ ${x}^{{2}}+{y}^{{2}}-{2}{h}{x}+{\left(-{6}{h}+{14}\right)}{y}+{\left({8}{h}-{36}\right)}={0}$ 1A  (b) Denote ${C}$ be the circle which passes through ${P}$ , ${Q}$ and ${R}$ . Note that the centre of ${C}$ lies on the perpendicular bisector of ${P}{Q}$ . Let ${h}$ be the ${x}$-coordinate of the centre of ${C}$ . By (a)(ii), after subsituting ${\left(-{25},-{1}\right)}$ , we have  ${\left(-{25}\right)}^{{2}}+{\left(-{1}\right)}^{{2}}-{2}{\left({h}\right)}{\left(-{25}\right)}+{\left(-{6}{h}+{14}\right)}{\left(-{1}\right)}+{\left({8}{h}-{36}\right)}={0}$ 1M for using (a)(ii) Solving, we have ${h}=-{9}$ . Hence, the equation of ${C}$ is ${x}^{{2}}+{y}^{{2}}+{18}{x}+{68}{y}-{108}$ $={0}$ . The required diameter $={2}\sqrt{{{\left(\dfrac{{18}}{{2}}\right)}^{{2}}+{\left(\dfrac{{68}}{{2}}\right)}^{{2}}-{\left(-{108}\right)}}}$ 1M $={2}\sqrt{{{1345}}}$ 1A

 (a)(i) The equation of ${L}$ is  ${\left({x}+{2}\right)}^{{2}}+{\left({y}-{2}\right)}^{{2}}$ $={\left({x}-{7}\right)}^{{2}}+{\left({y}+{1}\right)}^{{2}}$ 2M 1M+1M ${x}^{{2}}+{4}{x}+{4}+{y}^{{2}}-{4}{y}+{4}$ $={x}^{{2}}-{14}{x}+{49}+{y}^{{2}}+{2}{y}+{1}$ ${18}{x}-{6}{y}-{42}$ $={0}$ ${y}$ $={3}{x}-{7}$ 1A or equivalent

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