The circle through origin and cutting $x^{2} + y^{2} + 6 x - 15 = 0$ $x^{2} + y^{2} - 8 y + 10 = 0$ orthogonally is

2 x^{2} + 2 y^{2} - 10 x - 5 y = 0

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2 x^{2} + 2 y^{2} + 10 x + 5 y = 0

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x^{2} + y^{2} - 5 x + 5 y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 x^{2} + 2 y^{2} + 10 x - 5 y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $2 x^{2} + 2 y^{2} - 10 x - 5 y = 0$
If two circle having radii $r_{1} a n d r_{2}$ and centres at $c_{1} a n d c_{2}$ points respectively then
$c o s (180 - 90) = \frac{{r_{1}}^{2} + {r_{2}}^{2} - (c_{1} c_{2})^{2}}{2 r_{1} r_{2}} = c o s (90^{\circ})$
${r_{1}}^{2} + r_{2}^{2} = (c_{1} c_{2})^{2} \Rightarrow g_{1} g_{2} + f_{1} f_{2} = \frac{c_{1} + c_{2}}{2}$
Let $x^{2} + y^{2} + 2 g x + 2 f y = 0$ is a circle passing through $(0, 0)$
So, $2 g (3) + 2 f (0) = - 15$
$g =^{- 5} /_{2}$
and $2 g (0) + 2 f (- 4) = 10$
$f =^{- 5} /_{4}$
So, $e q^{n}$ of circle is
$x^{2} + y^{2} - 5 x - \frac{5}{2 y} = 0$
$\Rightarrow 2 (x^{2} + y^{2}) - 10 x - 5 y = 0$

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The circle through origin and cutting x2+y2+6x−15=0 x2+y2−8y+10=0 orthogonally is

The circle through origin and cutting $x^{2} + y^{2} + 6 x - 15 = 0$ $x^{2} + y^{2} - 8 y + 10 = 0$ orthogonally is