Question

If the circumcentre of the triangle whose vertices are $(0,2),(3,5)$ and $(5,8)$ is $(h,k)$ then $4(h^2+k^2)$ is equal to

Answer 1

Since $h,k$ is the centre of the circle, hence it is equidistant from the vertices of the triangle.
Therefore
$(h-0{)}^2+(k-2{)}^2=(h-3{)}^2+(k-5{)}^2$
$3(2h-3)+3(2k-7)=0$
$2h+2k=10$
$h+k=5$ ...(i)
$(h-5{)}^2+(k-8{)}^2=h^2+(k-2{)}^2$
$5(2h-5)+6(2k-10)=0$
$10h+12k=85$ ...(ii)
Solving i and ii, we get
$k=\frac{35}{2}$ and $h=\frac{-25}{2}$.
Hence
$4(h^2+k^2)$
$=4\left(\frac{{35}^2+{25}^2}{4}\right)$
$=1225+625$
$=1850$.