Question

A ( -3 ,4 ) , B ( -5 , 0 ) , C ( 3 , 0 ) are the vertices of triangle ABC . Find the coordinates of the circumcenter of triangle ABC . Also find the radius of circumcircle​

Answer 1

Let the circumcentre be O(h,k)

O is equidistant from all points

The points given are A(-3, 4) , B(-5, 0) and C(3 ,0)

The circumcircle passes through A, B and C

thus,

AO = BO = CO

==> AO² = BO² = CO²

Distance formula

D = $\sqrt{( {x_2 - x_1)}^{2} + (y_2 - y_1) {}^{2} }$

AO² = BO²

$\scriptsize{ \left(h -( - 3) \right) }^{2} + ( {k - 4)} ^{2} = \left( {h - ( - 5)} \right)^{2} + ( {k - 0)}^{2} \\ \\ \small(h + 3) {}^{2} + (k - 4) {}^{2} = {(h + 5)}^{2} + {k}^{2} \\ \\ \texttt{Using identities} \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ \\ \scriptsize {h}^{2} + 9 + 6h + {k}^{2} + 16 - 8k = {h}^{2} + 25 + 10h + {k}^{2} \\ \\ \scriptsize \cancel{h {}^{2} } + 6h + \cancel{k {}^{2} } + 9 + 16 - 8k = \cancel{h {}^{2} } + 25 + 10h + \cancel{k {}^{2} } \\ \\ \small 6h + 25 - 8k = 25 + 10h \\ \\ 6h - 8k + \cancel{25} = \cancel{25} + 10h \\ \\ - 8k = 10h - 6h \\ \\ - 8k = 4h \\ \\ k = - \frac{4h}{8}$

AO² = CO²

$\scriptsize \left(h - ( - 3) \right) {}^{2} + ( {k - 4)}^{2} = ( {h - 3)}^{2} + (k - 0) {}^{2} \\ \\ \small ( {h + 3)}^{2} + ( {k - 4)}^{2} = ( {h - 3)}^{2} + {k}^{2} \\ \\ \scriptsize {h}^{2} + 9 + 6h + {k}^{2} + 16 - 8k = {h}^{2} + 9 - 6h +{k}^{2} \\ \\ \scriptsize\cancel {h {}^{2} } + 6h + \cancel{k {}^{2} } + \cancel 9 + 16 - 8k = \cancel{h {}^{2} } + \cancel9 - 6h + \cancel{k {}^{2} } \\ \\ \small 16 - 8k = - 6h - 6h \\ \\ 16 - 8k = - 12h \\ \\ \texttt{Putting the value of k} \\ \\ 16 - 8 \left( - \frac{4h}{8} \right) = - 12h \\ \\ 16 - \cancel8 \left( - \frac{4h}{ \cancel8} \right) = - 12h \\ \\ 16 + 4h = - 12h \\ \\ 16 = - 12h - 4h \\ \\ 16 = - 16h \\ \\ h = - \frac{16}{16} \\ \\ \boxed { \green{h = - 1}}$

Now substituting the value of h in k = -4h/8

$k = - \frac{4( - 1)}{8} \\ \\ k = \frac{4}{8} \\ \\ \boxed{ \green{k = \frac{1}{2} }}$

Hence, the coordinates of circumcentre are $\boxed{O \left( - 1, \frac{1}{2} \right)}$

Radius = OC

$OC = \sqrt{( {x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} }$

$= \sqrt{ { \left(3 - ( - 1) \right)}^{2} + {(0 - \left( \frac{1}{2} \right) })^{2} } \\ \\ = \sqrt{ {(3 + 1)}^{2} + \left( \frac{1}{2} \right) {}^{2} } \\ \\ = \sqrt{ {4}^{2} + \frac{1}{4} } \\ \\ = \sqrt{16 + \frac{1}{4} } \\ \\ = \sqrt{ \frac{16(4) + 1}{4} } \\ \\ = \sqrt{ \frac{64 + 1}{4} } \\ \\ = \sqrt{ \frac{65}{4} } \\ \\ = \frac{ \sqrt{65} }{2}$

Hope it helps you