The line 2y=x+3 meets the circle x^2 +y^2 −2x−6y−15 at point M and N where N is in the second quadrant. find the coordinate of M and N. Mastermind


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Question Number 170378 by Mastermind last updated on 22/May/22

$T h e l i n e 2 y = x + 3 m e e t s t h e c i r c l e$ $x^{2} + y^{2} - 2 x - 6 y - 15 a t p o i n t M a n d N$ $w h e r e N i s i n t h e s e c o n d q u a d r a n t .$ $f i n d t h e c o o r d i n a t e o f M a n d N .$ $M a s t e r m i n d$
	Answered by daus last updated on 22/May/22

	$x^{2} + y^{2} - 2 x - 6 y = 15$ $2 y - 3 = x$ ${(2 y - 3)}^{2} + y^{2} - 2 (2 y - 3) - 6 y = 15$ $4 y^{2} - 12 y + 9 + y^{2} - 4 y + 6 - 6 y = 15$ $5 y^{2} - 22 y = 0$ $y (5 y - 22) = 0$ $y = 0, 5 y - 22 = 0$ $y = \frac{22}{5} o r 4.4$ $x = - 3, x = 5.8$ $N (- 3, 0) M (5.8, 4.4)$
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