Question

A circle touches both the coordinate axes and the line $x-y=a\sqrt{2}(a>0)$. The coordinates of the center of the circle can be

• A. $(a,a)$
• B. $(a,-a)$
• C. $(-a,a)$
• D. none of these

Answer 1

The correct option is A $(a,a)$

Let $r$ be the radius of the circle. Since it touches the coordinate axes and the line $x-y=a\sqrt{2}$ the coordinates of the centre of the circle can be $(r,r),(-r,-r)$ or $(r,-r)(Asr>0$ and the line $x-y=a\sqrt{2}$ meets the coordinates axes at $\left(a\sqrt{2}0\right)$ and $(0,-a\sqrt{2})).$
If the centre is $(r;r)$ or $(-r,-r)$ then $\left|\frac{-a\sqrt{2}}{\sqrt{1+1}}\right|=r\Rightarrow r=a$
So $(a,a)$ can be the coordinates of the centre of the circle, check that if the centre is $(r,-r)$
we have $\left|\frac{r+r-a\sqrt{2}}{\sqrt{2}}\right|=r\Rightarrow r=(\sqrt{2}\pm 1)a$