Question

The equation of a circle is $x^2+y^2=4$. Find the centre of the smallest circle touching the circle and the line $x+y=5\sqrt{2}$.

Answer 1

The minimum distance between these two curves is along with the line $y=x$.

Therefore, the centre of the smallest circle lies on the line $y=x$.

points on the smallest circle which are touching curves are

$(\sqrt{2},\sqrt{2})$ and $(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}})$

and the line joining these points is the diameter of the circle.

$\therefore$ centre of the circle is $(\frac{7}{2\sqrt{2}},\frac{7}{2\sqrt{2}})$