If the lines 2x+3y+1=0 and 3x-y-4=0 lie along diameters of a circle of circumference 10 $π$ , then the equation of the circle is

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$x^{2} + y^{2} - 2 x + 2 y - 23 = 0$

$x^{2} + y^{2} + 2 x + 2 y - 23 = 0$

$x^{2} + y^{2} + 2 x - 2 y - 23 = 0$

$x^{2} + y^{2} - 2 x - 2 y - 23 = 0$

answer is A.

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Detailed Solution

Point of intersection of 2x+3y+1=0, 3x-y-4=0 is (1, -1)

$\Rightarrow$ Centre of the circle= $(1, - 1)$

Circumference = $10 π \Rightarrow 2 π r = 10 π \Rightarrow r = 5 .$

Equation of the circle is ${(x - 1)}^{2} + {(y + 1)}^{2} = 5^{2} \Rightarrow x^{2} + y^{2} - 2 x + 2 y - 23 = 0 .$

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