let the cercle (x+1)^(2 ) +(y−3)^2 =9 and the point A(4,1) vrrify that A is out of circle and determine the equation of two tangentes to circle wich passes by point A.


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Question Number 77367 by msup trace by abdo last updated on 05/Jan/20

$l e t t h e c e r c l e {(x + 1)}^{2} + {(y - 3)}^{2} = 9$ $a n d t h e p o i n t A (4, 1)$ $v r r i f y t h a t A i s o u t o f c i r c l e$ $a n d d e t e r m i n e t h e e q u a t i o n o f$ $t w o t a n g e n t e s t o c i r c l e w i c h$ $p a s s e s b y p o i n t A .$
Commented by mathmax by abdo last updated on 05/Jan/20

$t h e p l a n i s p r o v i d e d w i t h o r t h o n o r m a l r e f e r e n c e (o, \vec{i}, \overset{\to)}{j}) .$
	Answered by jagoll last updated on 06/Jan/20

	$test point A (4, 1) \Rightarrow 5^{2} + {(- 2)}^{2} > 9$ $then A is out the circle$ $(2) let y = mx + n is tangent the circle$ $substitute point A \Rightarrow 1 = 4 m + n$ $n = 1 - 4 m . rewrite tangent line$ $y = mx + 1 - 4 m$ $distance of center circle to line equal$ $to radius \Rightarrow 3 =∣ \frac{- m + 1 - 4 m - 3}{\sqrt{1 + m^{2}}} ∣$ $3 \sqrt{1 + m^{2}} = ∣ - 5 m - 2 ∣$ $9 + {9 m}^{2} = {25 m}^{2} + 20 m + 4$ ${16 m}^{2} + 20 m - 5 = 0$ $m_{1} = \frac{1}{4} and m_{2} = - \frac{5}{4} . now you$ $get the tangent line .$
	Commented by msup trace by abdo last updated on 06/Jan/20

	$t h a n k s s i r .$
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