Question

A circle touches the line $y=x$ at a point $P$ such that $OP=4\sqrt{2}$, where $O$ is the origin. The circle makes an intercept of $6\sqrt{2}$ units on line $x+y=0.$ Then the equation of the circle(s) is/are

• A. $(x-9{)}^2+(y+1{)}^2-50=0$
• B. $(x-1{)}^2+(y+9{)}^2-50=0$
• C. $(x+1{)}^2+(y-9{)}^2-50=0$
• D. $(x+9{)}^2+(y-1{)}^2-50=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(x-9{)}^2+(y+1{)}^2-50=0$
B $(x-1{)}^2+(y+9{)}^2-50=0$
C $(x+1{)}^2+(y-9{)}^2-50=0$
D $(x+9{)}^2+(y-1{)}^2-50=0$

Let $AB$ be the length of chord intercepted by circle on $y+x=0$
Let $CM$ be perpendicular to $AB$ from centre $C(h,k)$.

Also, $y-x=0$ and $y+x=0$ are perpendicular to each other.
$\therefore OPCM$ is rectangle.
$\therefore CM=OP=4\sqrt{2}$.
and $AB=6\sqrt{2}$
Now $AB=2AM=2\sqrt{r^2-{\left(CM\right)}^2}$
$\Rightarrow r=5\sqrt{2}$

Again $y=x$ touches the circle.
$\therefore CP=r$
$\Rightarrow \left|\frac{h-k}{\sqrt{2}}\right|=5\sqrt{2}\Rightarrow h-k=\pm 10\dots \left(1\right)$
Also, $CM=4\sqrt{2}$
$\Rightarrow \left|\frac{h+k}{\sqrt{2}}\right|=4\sqrt{2}\Rightarrow h+k=\pm 8\dots \left(2\right)$
Solving equations $\left(1\right)$ and $\left(2\right)$, we get the possible centres as $(9,-1),(1,-9),(-1,9),(-9,1)$

$\therefore$ Possible circles are
$(x-9{)}^2+(y+1{)}^2-50=0$
$(x-1{)}^2+(y+9{)}^2-50=0$
$(x+1{)}^2+(y-9{)}^2-50=0$
$(x+9{)}^2+(y-1{)}^2-50=0$