Question

Find the equation of the circle which passes through the points $(5,0)$ and $(1,4)$ whose centre lies on the line $x+y-3=0$.

Answer 1

Let the centre be $(h,3-h)$, distance of centre from both points would be equal (radius)

$\Rightarrow \sqrt{(h-5{)}^2+(h-3{)}^2}=\sqrt{(h-1{)}^2+(h+1{)}^2}$

$\Rightarrow 2h^2-10h-6h+34$

$=2h^2-2h+2h+2$

$\Rightarrow -16h+32=0$

$\Rightarrow h=2$

$\Rightarrow$ Centre $=(2,1)$

Radius $=\sqrt{9+1}=\sqrt{10}$

$\Rightarrow$ Equation of circle $=(x-2{)}^2+(y-{)}^2=10$

$\Rightarrow x^2+y^2-4x-2y-5=0$.