Question

If the circle $x^2+y^2+2gx+2fy+c=0$ touches by the line $y=x$ at the point $P$ such that $OP=6\sqrt{2}$, where $O$ is the origin, then the value of $c$ is equal to

• A. 74
• B. 62
• C. 64
• D. 72

Answer 1

Answer: (D)

72

The equation of the line $y=x$ in parametric form is
$\frac{x}{\cos \pi /4}=\frac{y}{\sin \pi /4}$
$\because OP=6\sqrt{2}$
Therefore, coordinates of $P$ are given by,
$\frac{x}{\cos \frac{\pi }{4}}=\frac{y}{\cos \frac{\pi }{4}}$
$\Rightarrow x=y=6$
Thus coordinates of $P$ are $(6,6)$.
The equation of cirde touching $y=x$ at $P(6,6)$ is
$(x-6{)}^2+(y-6{)}^2+\lambda (x-y)=0$
$\Rightarrow x^2+y^2+x(\lambda +12)+(\lambda +12)+72=0$
Comparing it with $x^2+y^2+2gx+2fy+c=0$
$\lambda -12=2g,-(\lambda +12)=2f$ and $c=72$
Hence, required value of $c$ is $72$.