If the line $3 x + 2 λ y + 12 = 0$ is a diameter of the circle $x^{2} + y^{2} - 4 x + 6 y + 12 = 0,$ then the value of $λ$ is

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Solution

Given line is
$3 x + 2 λ y + 12 = 0$
Equation of the circle is
$x^{2} + y^{2} - 4 x + 6 y + 12 = 0$ ,
Centre of the circle is
$C = (- g, - f) = (2, - 3)$
Centre should lie on the equation of diameter, so
$3 (2) + 2 λ (- 3) + 12 = 0 ∴ λ = 3$

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3 x + 2 λ y + 12 = 0

x^{2} + y^{2} - 4 x + 6 y + 12 = 0,

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The diameters of a circle are along $2 x + y - 7 = 0$ and $x + 3 y - 11 = 0$ Then the equation of this circle, which also passes through (5,7) is

x^{2} + y^{2} + 2 λ x - 2 λ y + 18 = 0

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λ

S_{1} \equiv x^{2} + y^{2} - 4 x - 6 y - 12 = 0

S_{2} \equiv x^{2} + y^{2} + 6 x + 4 y - 12 = 0

L \equiv x + y = 0

S_{1} = x^{2} + y^{2} - 4 x - 6 y - 12 = 0

S_{2} = x^{2} + y^{2} + 6 x + 4 y - 12 = 0

L \equiv x + y = 0

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