Question

# Find the co-ordinates of the centre of a circle passing through the points A(3,5) B($-$1, 2) and C(3, $-$3). Also find the radius of circle.

Open in App
Solution

## $\mathrm{Let}\mathrm{the}\mathrm{coordinate}\text{s}\mathrm{of}\mathrm{circumcentre}\mathrm{O}\mathrm{be}\left(\mathrm{x},\mathrm{y}\right).\phantom{\rule{0ex}{0ex}}\mathrm{Distance}\mathrm{of}\mathrm{circumcentre}\mathrm{from}\mathrm{the}\mathrm{vertices}\mathrm{A}\left(3,5\right),\mathrm{B}\left(-1,2\right)\mathrm{and}\mathrm{R}\left(3,-3\right)\mathrm{of}\mathrm{the}\mathrm{triangle}\phantom{\rule{0ex}{0ex}}\mathrm{will}\mathrm{be}\mathrm{equal}\mathrm{as}\mathrm{it}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}.$ $\therefore \mathrm{OA}=\mathrm{OB}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{OA}}^{2}={\mathrm{OB}}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-3\right)}^{2}+{\left(y-5\right)}^{2}={\left[x-\left(-1\right)\right]}^{2}+{\left(y-2\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-3\right)}^{2}+{\left(y-5\right)}^{2}={\left(x+1\right)}^{2}+{\left(y-2\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+9-6x+{y}^{2}+25-10y={x}^{2}+1+2x+{y}^{2}+4-4y\phantom{\rule{0ex}{0ex}}⇒-6x+34-10y=2x+5-4y\phantom{\rule{0ex}{0ex}}⇒-8x-6y=-29\phantom{\rule{0ex}{0ex}}⇒8x+6y=29----\left(1\right)\phantom{\rule{0ex}{0ex}}$ $\therefore \mathrm{OB}=\mathrm{OC}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{OB}}^{2}={\mathrm{OC}}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left[x-\left(-1\right)\right]}^{2}+{\left(y-2\right)}^{2}={\left(x-3\right)}^{2}+{\left[y-\left(-3\right)\right]}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x+1\right)}^{2}+{\left(y-2\right)}^{2}={\left(x-3\right)}^{2}+{\left(y+3\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+1+2x+{y}^{2}+4-4y={x}^{2}+9-6x+{y}^{2}+9+6y\phantom{\rule{0ex}{0ex}}⇒2x-4y+5=-6x+6y+18\phantom{\rule{0ex}{0ex}}⇒8x-10y=13----\left(2\right)\phantom{\rule{0ex}{0ex}}$ $\mathrm{Subtracting}\mathrm{equation}\left(2\right)\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}8x+6y=29\phantom{\rule{0ex}{0ex}}8x-10y=13\phantom{\rule{0ex}{0ex}}\overline{)-+-}\phantom{\rule{0ex}{0ex}}16y=16\phantom{\rule{0ex}{0ex}}⇒y=1\phantom{\rule{0ex}{0ex}}\mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}y\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}8x+6×1=29\phantom{\rule{0ex}{0ex}}⇒8x=23\phantom{\rule{0ex}{0ex}}⇒y=\frac{23}{8}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\left(\frac{23}{8},1\right)\text{are}\mathrm{the}\mathrm{coordinate}s\mathrm{of}\mathrm{the}\mathrm{circumcentre}\mathrm{of}\mathrm{triangle}\mathrm{PQR}$ $\mathrm{Now},\mathrm{radius}=\mathrm{OA}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{23}{8}-3\right)}^{2}+{\left(1-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{-1}{8}\right)}^{2}+{\left(-4\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{64}+16}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1025}{64}}\phantom{\rule{0ex}{0ex}}=\frac{5\sqrt{41}}{8}\mathrm{units}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Suggest Corrections
0
Related Videos
Finding the Coordinates of a Point
MATHEMATICS
Watch in App