Question

If the circle $x^2+y^2+2gx+2fy+c=0$ is touched by y = x at P in the first quadrant, such that $OP=6\sqrt{2}$, then the value of $c$ is

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

Answer 1

Answer: (C)

72

Let the point of contact be $x_1,y_1$
However it lies on the line $y=x$
Hence
$x_1=y_1$
Applying distance formula, we get
$\sqrt{x_1^2+x_1^2}=6\sqrt{2}$
$2x_1^2=72$
$x_1=6$ ...(positive, since it lies on y=x.)
Differentiating the equation of circle with respect to x.
$2x+2y{y}^{\prime }+2g+2f{y}^{\prime }=0$
Now $y^{\prime }=1$ since $y^{\prime }$ is the slope of the line $y=x$.
By substituting, we get
$x+y=-(g+f)$
Now $x_1=y_1=6$
Hence
$12=-(g+f)$ ...(i)
Substituting $x=y=6$ in the equation of the circle, we get
$36+36+2\left(6\right)(g+f)+c=0$
$72+12(-12)+c=0$
$c=144-72$
$c=72$