Question

If $4x+3y-12=0$ touches ${\left(x-p\right)}^2+{\left(y-p\right)}^2=p^2$ , then the sum of all the possible values of $p$ is

Answer 1

Answer: 7

The distance of the centre $\left(p,p\right)$ of the given circle from the line $4x+3y-12=0$ is equal to the radius $\left(\left|p\right|\right)$
$\Rightarrow \left|\frac{4p+3p-12}{5}\right|=\left|p\right|$
$\Rightarrow 7p-12=5p$ or $7p-12=-5p$
$\Rightarrow p=6$ or $p=1$
$\Rightarrow$ sum of all the possible values of $‘p’=6+1=7$