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Question Number 138548 by I want to learn more last updated on 14/Apr/21

	Answered by nimnim last updated on 14/Apr/21

	$L e t t h e c e n t r e o f t h e c i r c l e b e C (h, k)$ $S i n c e, i t t o u c h e s t h e x - a x i s a t P (- 3, 0),$ $C P ⊥ x - a x i s a n d h = - 3$ $s i n c e, t h e c e n t r e l i e s o n t h e l i n e 2 x + y + 1 = 0$ $\Rightarrow 2 (- 3) + k + 1 = 0 \Rightarrow k = 5$ $(a) c o o r d i n a t e o f C = (- 3, 5)$ $(b) r a d i u s = 5 u n i t s (∵ C P ⊥ x - a x i s)$ $(c) e q u a t i o n o f t h e c i r c l e :$ ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $\Rightarrow {(x + 3)}^{2} + {(y - 5)}^{2} = 5^{2}$ $\Rightarrow x^{2} + 6 x + 9 + y^{2} - 10 y + 25 = 25$ $\Rightarrow x^{2} + y^{2} + 6 x - 10 y + 9 = 0$
	Commented by I want to learn more last updated on 14/Apr/21

	$Thanks sir, i appreciate .$
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