JEE Main 2018 — Circle Question with Solution

The equation of a tangent to the given circle at $\left(2,2\right)$ is

$x\left(2\right)+y\left(2\right)+\left(x+2\right)-2\left(y+2\right)-4=0\Rightarrow 3x-6=0\Rightarrow x=2$

Family of the circle touching a given line at a given point

${\left(x-2\right)}^2+{\left(y-2\right)}^2+\lambda \left(x-2\right)=0$.

$\Rightarrow x^2+y^2+\left(\lambda -4\right)x-4y+\left(8-2\lambda \right)=0$

Radius$=3=\sqrt{{\left(\frac{\lambda -4}{2}\right)}^2+4-\left(8-2\lambda \right)}$

$\Rightarrow 9=\frac{{\lambda }^2-8\lambda +16}{4}-4+2\lambda$

$\Rightarrow 36={\lambda }^2-8\lambda +16-16+8\lambda$

$\Rightarrow \lambda =\pm 6$

Hence, equation of the circle is $x^2+y^2-10x-4y+20=0$.

So, $x$-intercept$=2\sqrt{25-20}=2\sqrt{5}$.