Find the equation of the circle which is orthogonal to the circles x^2 +y^2 −7x−y=0 and x^2 +y^2 +3x−6y+5=0 and which passes through the point (−3,0).


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Question Number 142208 by ZiYangLee last updated on 27/May/21

$Find the equation of the circle which is$ $orthogonal to the circles x^{2} + y^{2} - 7 x - y = 0$ $and x^{2} + y^{2} + 3 x - 6 y + 5 = 0 and which passes$ $through the point (- 3, 0) .$
	Answered by mr W last updated on 27/May/21

	$x^{2} + y^{2} - 7 x - y = 0$ $\Rightarrow {(x - \frac{7}{2})}^{2} + {(y - \frac{1}{2})}^{2} = {(\frac{5}{\sqrt{2}})}^{2}$ $\Rightarrow c e n t e r (\frac{7}{2}, \frac{1}{2}), r a d i u s \frac{5}{\sqrt{2}}$ $x^{2} + y^{2} + 3 x - 6 y + 5 = 0$ ${(x + \frac{3}{2})}^{2} + {(y - 3)}^{2} = {(\frac{5}{2})}^{2}$ $\Rightarrow c e n t e r (- \frac{3}{2}, 3), r a d i u s \frac{5}{2}$ $s a y t h e e q u a t i o n o f t h e t h i r d c i r c l e i s$ ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ ${(\frac{7}{2} - a)}^{2} + {(\frac{1}{2} - b)}^{2} = {(r + \frac{5}{\sqrt{2}})}^{2} . . . (i)$ ${(- \frac{3}{2} - a)}^{2} + {(3 - b)}^{2} = {(r + \frac{5}{2})}^{2} . . . (i i)$ ${(- 3 - a)}^{2} + {(0 - b)}^{2} = r^{2} . . . (i i i)$ $(i) - (i i i) :$ $(\frac{1}{2} - 2 a) (\frac{13}{2}) + (\frac{1}{2} - 2 b) (\frac{1}{2}) = \frac{5}{\sqrt{2}} (2 r + \frac{5}{\sqrt{2}})$ $26 a + 2 b = - 10 \sqrt{2} r - 18 . . . (I)$ $(i i) - (i i i) :$ $(- \frac{9}{2} - 2 a) (\frac{3}{2}) + (3 - 2 b) (3) = \frac{5}{2} (2 r + \frac{5}{2})$ $3 a + 6 b = - 5 r - 4 . . . (I I)$ $\Rightarrow a = - \frac{(6 \sqrt{2} - 1) r}{15} - \frac{2}{3}$ $\Rightarrow b = \frac{(3 \sqrt{2} - 13) r}{15} - \frac{1}{3}$ $p u t i n t o (i i i) :$ ${(\frac{(6 \sqrt{2} - 1) r}{15} - \frac{7}{3})}^{2} + {(\frac{(3 \sqrt{2} - 13) r}{15} - \frac{1}{3})}^{2} = r^{2}$ $(18 \sqrt{2} - 7) r^{2} + (90 \sqrt{2} - 40) r - 250 = 0$ $r = \frac{- 45 \sqrt{2} + 20 \pm 30 \sqrt{3 + 3 \sqrt{2}}}{18 \sqrt{2} - 7}$ $(n e g a t i v e v a l u e f o r t h e b i g o n e)$
	Commented by mr W last updated on 27/May/21

	Commented by mr W last updated on 27/May/21

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