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 contains the point $(-10,2)$ in its interior and the length of its chord on the line $x+y=0$ is $6\sqrt{2}$. The equation of the circle is

• A. $x^2+y^2+18x-2y+32=0$
• B. $x^2+y^2-18x-2y+32=0$
• C. $x^2+y^2+18x+2y+32=0$
• D. None of these

Answer 1

Answer: (A)

$x^2+y^2+18x-2y+32=0$

Let $C(\alpha ,\beta )$ be the center of the circle touching $OP$ at $P$ and making intercept $AB=6\sqrt{2}$ on the line $x+y=0$ as shown in the figure.

If $r$ is the radius of the circle, then

$r^2={AC}^2={CL}^2+{LA}^2={\left(\frac{\alpha +\beta }{\sqrt{2}}\right)}^2+{\left(3\sqrt{2}\right)}^2$ ...(1)

Since $OP=4\sqrt{2}$

$\therefore {OC}^2={CP}^2+{PO}^2\Rightarrow {\alpha }^2+{\beta }^2=r^2+{\left(4\sqrt{2}\right)}^2$ ...(2)

From (1) and (2), we get

${\alpha }^2+{\beta }^2={\left(\frac{\alpha +\beta }{\sqrt{2}}\right)}^2+18+32$

$\Rightarrow ({\alpha }^2+{\beta }^2)-\frac{{(\alpha +\beta )}^2}{2}=50\Rightarrow {(\alpha -\beta )}^2=100$$\Rightarrow \alpha -\beta =\pm 10$ ...(3)

Also, $CP=r\Rightarrow r=\left|\frac{\alpha -\beta }{\sqrt{2}}\right|\Rightarrow r^2=\frac{{(\alpha -\beta )}^2}{3}$ ...(4)

From (3), and (4), we get

$r^2={\left(5\sqrt{2}\right)}^2\Rightarrow r=5\sqrt{2}.$

Substituting $r=5\sqrt{2}$ in (1), we get

${\left(5\sqrt{2}\right)}^2={\left(\frac{\alpha +\beta }{\sqrt{2}}\right)}^2+18$

$\Rightarrow 32=\frac{{(\alpha +\beta )}^2}{2}\Rightarrow {(\alpha +\beta )}^2=64$$\Rightarrow \alpha +\beta =\pm 8$ ...(5)

From (3), and (5), we get

$\alpha =9,\beta =-1$ or $\alpha =-9,\beta =1$

or $\alpha =1,\beta =-9$ or $\alpha =-1,\beta =9.$

Since $Q(-10,2)$ lies in the interior of the circle,

$\therefore CQ$ must be less than $5\sqrt{2}.$

Thus, centre of the circle must be $(-9,1)$

Therefore, the equation of the required circle is

${(x+9)}^2+{(y-1)}^2={\left(5\sqrt{2}\right)}^2$$\Rightarrow x^2+y^2+18x-2y+32=0.$