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-

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} \dots (i)

{(- \frac{3}{2} - a)}^{2} + {(3 - b)}^{2} = {(r + \frac{5}{2})}^{2} \dots (i i)

{(- 3 - a)}^{2} + {(0 - b)}^{2} = r^{2} \dots (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 \dots (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 \dots (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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