if-the-circle-x-2-y-2-2y-8-0-and-x-2-y-2-24x-hy-0-cut-orthogonally-determine-the-value-of-h-

Question Number 37591 by mondodotto@gmail.com last updated on 15/Jun/18

if the circle x^{2} + y^{2} - 2 y - 8 = 0

and x^{2} + y^{2} - 24 x + h y = 0 cut orthogonally,

determine the value of h .

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jun/18

x^{2} + y^{2} - 2 y - 8 = 0

c o m p a r i n g w i t h x^{2} + y^{2} + 2 g x + 2 f y + c = 0

c e n t r e (- g, - f) r a d i u s = \sqrt{g^{2} + f^{2} - c}

o_{1} (0, 1) r_{1} = \sqrt{0^{2} + {(- 1)}^{2} - (- 8)} = \sqrt{9} = 3

o_{2} (12, - \frac{h}{2}) r_{2} = \sqrt{{(- 12)}^{2} + {(\frac{h}{2})}^{2}}

r_{2} = \sqrt{144 + \frac{h^{2}}{4}}

o_{1} o_{2} = \sqrt{{(12 - 0)}^{2} + {(- \frac{h}{2} - 1)}^{2}}

= \sqrt{144 + \frac{h^{2}}{4} + h + 1} = \sqrt{\frac{h^{2}}{4} + h + 145}

r_{1}^{2} + r_{2}^{2} = {(o_{1} o_{2})}^{2}

9 + 144 + \frac{h^{2}}{4} = \frac{h^{2}}{4} + h + 145

h = 8

o r m d t h o d

c l n d i t i o n f o r o r t h o g o n s l i t y

2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}

2 \times 0 \times (- 12) + 2 \times (- 1) \times \frac{h}{2} = - 8 + 0

- h = - 8

h = 8