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Equation of Family of circles passing through point of intersection of two circles

The equation ...

Question

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

A

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

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B

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

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C

x^{2} + y^{2} - 3 x - 4 y = 0

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D

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

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Solution

The correct option is B $x^{2} + y^{2} + 4 x - 2 y - 4 = 0$
Required equation of the circle,
$(x^{2} + y^{2} - 2 x - 6 y + 6) + λ (3 x + 2 y - 5) = 0,$
This circle passing through points (-2, 4), therefore (4 + 16 + 4 - 24 + 6) + $λ$ (-6 + 8 - 5) = 0, $∴ λ$ = 2

$∴ (x^{2} + y^{2} - 2 x - 6 y + 6) + 2 (3 x + 2 y - 5) = 0$
$\Rightarrow x^{2} + y^{2} + 4 x - 2 y - 4 = 0.$

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Similar questions

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