Question

Prove that the centre of the smallest circle passing through origin and whose centre lies on $y=x+1$ is $\left(\begin{array}{c}-\frac{1}{2},\frac{1}{2}\end{array}\right)$

Answer 1

Let centre be $C(h,h+1)$, and $O(0,0)$ is the origin.

$r=OC=\sqrt{h^2+(h+1{)}^2}=\sqrt{2h^2+2h+1}$

$=\sqrt{2\left(\begin{array}{c}{h}^2+2\cdot \frac{1}{2}\cdot x+\frac{1}{4}\end{array}\right)+\frac{1}{2}}$

$=\sqrt{2{\left(\begin{array}{c}h+\frac{1}{2}\end{array}\right)}^2+\frac{1}{2}}$

For min radius r, $h+\frac{1}{2}=0,h=-\frac{1}{2}$

Thus, centre $(h,h+1)=\left(\begin{array}{c}-\frac{1}{2},\frac{1}{2}\end{array}\right)$