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

x^{2} + y^{2} + 2 (x + y) - 23 = 0

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x^{2} + y^{2} - 2 (x + y) - 23 = 0

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

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

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Solution

The correct option is C $x^{2} + y^{2} - 2 x + 2 y - 23 = 0$
Given lines $2 x + 3 y + 1 = 0$ and $3 x - y - 4 = 0$
Point of intersection of these lines is the center of circle.
So, center is at (1,-1).
Also, given circumference $= 10 π$
$\Rightarrow 2 π r = 10 π$
$\Rightarrow r = 5$
Hence the equation of circle is
$(x - 1)^{2} + (y + 1)^{2} = 25$
$x^{2} + y^{2} - 2 x + 2 y - 23 = 0$

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