Find the equation of the circle which passes through the centre of the circle $x^{2} + y^{2} + 8 x + 10 y - 7 = 0$ and is concentric with the circle $2 x^{2} + 2 y^{2} - 8 x - 12 y - 9 = 0.$

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Solution

We have $2 x^{2} + 2 y^{2} - 8 x - 12 y - 9 = 0$

$\Rightarrow x^{2} + y^{2} - 4 x - 6 y - \frac{9}{2} = 0 \Rightarrow x^{2} + y^{2} + 2 g x + 2 f y + c = 0,$

where $g = - 2, f = - 3$ and $c = \frac{- 9}{2}$

Centre of this circle = $(- g, - f) = (2, 3) .$

$∴$ the centre of the required circle = $C (2, 3) .$

Again, $x^{2} + y^{2} + 8 x + 10 y - 7 = 0$

$\Rightarrow x^{2} + y^{2} + 2 g x + 2 f y + c = 0,$ where $g = 4, f = 5$ and $c = - 7.$

Centre of this circle = $(- g, - f) = (- 4, - 5) .$

So, the required circle passes through the point $P (- 4, - 5) .$

Radius of the required circle = $C P = \sqrt{(2 + 4)^{2} + (3 + 5)^{2}}$

= $\sqrt{36 + 64} = \sqrt{100} = 10$

Hence, the required equation of the circle is,

$(x - 2)^{2} + (y - 3)^{2} = (10)^{2} \Rightarrow x^{2} + y^{2} - 4 x - 6 y - 87 = 0.$

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Similar questions

x^{2} + y^{2} + 8 x + 10 y - 7 = 0

2 x^{2} + 2 y^{2} - 8 x - 12 y - 9 = 10

x^{2} + y^{2} + 8 x + 10 y - 7 = 0

2 x^{2} + 2 y^{2} - 8 x - 12 y - 9 = 0

x^{2} + y^{2} + 8 x + 10 y - 7 = 0

x^{2} + y^{2} - 4 x - 6 y = 0

x^{2} + y^{2} + 8 x + 10 y - 7 = 0

x^{2} + y^{2} - 4 x - 6 y = 0

x^{2} + y^{2} - 8 x - 12 y + 15 = 0

(5, 4)

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