Question

The coordinates of the centre of the smallest circle touching the circle $x^2+y^2=4$ and the lines $x+y=5\sqrt{2}$ are

• A. $\left(\frac{7}{2\sqrt{2}},\frac{7}{2\sqrt{2}}\right)$
• B. $\left(\frac{3}{2},\frac{3}{2}\right)$
• C. $\left(\frac{7}{2\sqrt{2}},\frac{3}{2}\right)$
• D. $(-\frac{3}{2},-\frac{3}{2})$

Answer 1

The correct option is A $\left(\frac{7}{2\sqrt{2}},\frac{7}{2\sqrt{2}}\right)$

By Geometry, $y=x$ is a equation of diameter of small circle
so, A $(a,a)$ is a center of small circle.
perpendicular distance of $(0,0)$ for $x+y=5\sqrt{2}$ is
$d=2r+2$
where, $r=$radius of small circle.
$d=2r+2=\left|\frac{5\sqrt{2}}{\sqrt{2}}\right|$
$2r=3$
$\Rightarrow r=\frac{3}{2}$
$\therefore a=(2+r)\cos {45}^0$
$a=\frac{7}{2}\times \frac{1}{\sqrt{2}}$
so, center of small circle is $(\frac{7}{2\sqrt{2}},\frac{7}{2\sqrt{2}})$