Find the equation of the circle which passes | vedclass

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(A) The given circle is $x^{2} + y^{2} - 8x + 6y - 5 = 0$. Comparing this with $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we get $g = -4$ and $f = 3$. The ce

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$x^{2} + y^{2} - 8x + 6y - 27 = 0$

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(A) The given circle is $x^{2} + y^{2} - 8x + 6y - 5 = 0$. Comparing this with $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we get $g = -4$ and $f = 3$. The center of the circle is $(-g, -f) = (4, -3)$. Since the required circle is concentric with the given circle,its center is also $(4, -3)$. The circle passes through the point $(-2, -7)$. The radius $r$ is the distance between $(4, -3)$ and $(-2, -7)$: $r^{2} = (4 - (-2))^{2} + (-3 - (-7))^{2} = (6)^{2} + (4)^{2} = 36 + 16 = 52$. The equation of the circle is $(x - 4)^{2} + (y + 3)^{2} = 52$. Expanding this,we get $x^{2} - 8x + 16 + y^{2} + 6y + 9 = 52$,which simplifies to $x^{2} + y^{2} - 8x + 6y - 27 = 0$.

Find the equation of the circle which passes through the point $(-2, -7)$ and is concentric with the circle $x^{2} + y^{2} - 8x + 6y - 5 = 0$.

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