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(B) The given circle is $x^2 + y^2 + 8x + 10y - 7 = 0$. Any circle concentric with this circle is of the form $x^2 + y^2 + 8x + 10y + k = 0$. The cent

Vedclass · Accepted Answer

$x^2 + y^2 + 8x + 10y - 59 = 0$

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(B) The given circle is $x^2 + y^2 + 8x + 10y - 7 = 0$. Any circle concentric with this circle is of the form $x^2 + y^2 + 8x + 10y + k = 0$. The centre of the circle $x^2 + y^2 - 4x - 6y = 0$ is found by comparing it with $x^2 + y^2 + 2gx + 2fy + c = 0$,which gives $(-g, -f) = (2, 3)$. Since the required circle passes through $(2, 3)$,we substitute these coordinates into the equation: $(2)^2 + (3)^2 + 8(2) + 10(3) + k = 0$ $4 + 9 + 16 + 30 + k = 0$ $59 + k = 0 \implies k = -59$. Thus,the equation of the circle is $x^2 + y^2 + 8x + 10y - 59 = 0$.

The equation of the circle concentric with the circle $x^2 + y^2 + 8x + 10y - 7 = 0$ and passing through the centre of the circle $x^2 + y^2 - 4x - 6y = 0$ is

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