If a circle touches the lines 3x - 4y - 10 = | vedclass

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(A) The distance between the parallel lines $3x - 4y - 10 = 0$ and $3x - 4y + 30 = 0$ is the diameter of the circle.Diameter $d = \frac{|30 - (-10)|}{

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$x^2 + y^2 + 4x - 2y - 11 = 0$

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(A) The distance between the parallel lines $3x - 4y - 10 = 0$ and $3x - 4y + 30 = 0$ is the diameter of the circle.Diameter $d = \frac{|30 - (-10)|}{\sqrt{3^2 + (-4)^2}} = \frac{40}{5} = 8$.Thus,the radius $r = \frac{d}{2} = 4$.The centre $(h, k)$ lies on the line $x + 2y = 0$,so $h = -2k$.The centre is equidistant from the two parallel lines. The line $3x - 4y + c = 0$ midway between the given lines is $3x - 4y + 10 = 0$.Substituting $x = -2k$ into $3x - 4y + 10 = 0$ gives $3(-2k) - 4k + 10 = 0$,which simplifies to $-10k = -10$,so $k = 1$.Then $h = -2(1) = -2$. The centre is $(-2, 1)$.The equation of the circle is $(x + 2)^2 + (y - 1)^2 = 4^2$.$x^2 + 4x + 4 + y^2 - 2y + 1 = 16$.$x^2 + y^2 + 4x - 2y - 11 = 0$.

If a circle touches the lines $3x - 4y - 10 = 0$ and $3x - 4y + 30 = 0$ and its centre lies on the line $x + 2y = 0$,then the equation of the circle is

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