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

Question

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

MARKS App · Accepted Answer

The distance of the centre p , p of the given circle from the line 4 x + 3 y - 12 = 0 is equal to the radius p ⇒ 4 p + 3 p - 12 5 = p ⇒ 7 p - 12 = 5 p or 7 p - 12 = - 5 p ⇒ p = 6 or p = 1 ⇒ sum of all the possible values of ‘ p ’ = 6 + 1 = 7

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