Question

If the circle $x^2+y^2+2gx+2fy+c=0$ is touched by $y=x$ at P such that $OP=6\sqrt{2}$, then the value of c is
($O$ is the origin of co-ordinate system)

• A. $36$
• B. $144$
• C. $72$
• D. None of these

Answer 1

The correct option is C $72$

The equation of the line $y=x$ in distance form is $\frac{x}{cos\theta }=\frac{y}{sin\theta }=r$, where $\theta =\frac{\pi }{4}$.
For point P, $r=6\sqrt{2}$.
Therefore, coordinates of P are given by $\frac{x}{cos\frac{\pi }{4}}=\frac{y}{sin\frac{\pi }{4}}=6\sqrt{2}\Rightarrow x=6,y=6$.
Since $P(6,6)$ lies on $x^2+y^2+2gx+2fy+c=0$,
Hence, $72+12(g+f)+c=0$ ......... (i)
Since, $y=x$ touches the circle, therefore the equation $2x^2+2x(g+f)+c=0$ has equal roots
$\Rightarrow 4(g+f{)}^2=8c$
$\Rightarrow (g+f{)}^2=2c$ ........ (ii)
From (i), we get
$\left[12\right(g+f){]}^2=[-(c+72){]}^2$
$\Rightarrow 144(g+f{)}^2=(c+72{)}^2$
$\Rightarrow 144\left(2c\right)=(c+72{)}^2$
$\Rightarrow (c-72{)}^2=0\Rightarrow c=72$
Hence, option 'C' is correct.