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Equation of Family of Circles Passing through Points of Intersection of Circle and a Line

The equation ...

Question

The equation of the circle passing through the point (1,2) and through the points of intersection of $x^{2} + y^{2} - 4 x - 6 y - 21 = 0$ and 3x+4y+5=0 is given by

x^{2} + y^{2} + 2 x + 2 y + 11 = 0

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

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x^{2} + y^{2} + 2 x - 2 y - 3 = 0

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x^{2} + y^{2} + 2 x + 2 y - 11 = 0

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Solution

The correct option is D $x^{2} + y^{2} + 2 x + 2 y - 11 = 0$
The family of circles passing through intersection of a given circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and given line $l x + m y + n = 0$ is,
$x^{2} + y^{2} + 2 g x + 2 f y + c + λ (l x + m y + n) = 0$ where $λ \in R$

So, for given case, we can write equation as,

$x^{2} + y^{2} - 4 x - 6 y - 21 + λ (3 x + 4 y + 5) = 0$

It is given that circle passes through (1,2), so
$1^{2} + 2^{2} - 4 - 12 - 21 + λ (3 + 8 + 5) = 0$
$\to 16 λ = 32$
$\to λ = 2$

So, equation of cirlce,
$\to x^{2} + y^{2} - 4 x - 6 y - 21 - 2 (3 x + 4 y + 5) = 0$
$\to x^{2} + y^{2} + 2 x + 2 y - 11 = 0$

Hence, (D) is the correct answer.

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The equation of the circle passing through the point (1,2) and through the points of intersection of x2+y2−4x−6y−21=0 and 3x+4y+5=0 is given by

The equation of the circle passing through the point (1,2) and through the points of intersection of $x^{2} + y^{2} - 4 x - 6 y - 21 = 0$ and 3x+4y+5=0 is given by