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  4. If 4x+3y-12=0 touches (x - p)2+(y - p)2=p2 , then the sum of all the possible values of p is

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Q. 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

NTA AbhyasNTA Abhyas 2020Conic Sections

Solution:

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 $\left(\left|p\right|\right)$
$\Rightarrow \left|\frac{4 p + 3 p - 12}{5}\right|=\left|p\right|$
$\Rightarrow 7p-12=5p$ or $7p-12=-5p$
$\Rightarrow p=6$ or $p=1$
$\Rightarrow $ sum of all the possible values of $‘p’=6+1=7$