If a system of circles passes through (2,3) a | vedclass

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(B) Let the centre of the circle be $(h, k)$. Since the circle passes through $(2, 3)$,its radius $r$ is given by $r^2 = (h-2)^2 + (k-3)^2$.The equati

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$4x + 6y - 25 = 0$

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(B) Let the centre of the circle be $(h, k)$. Since the circle passes through $(2, 3)$,its radius $r$ is given by $r^2 = (h-2)^2 + (k-3)^2$.The equation of the circle is $(x-h)^2 + (y-k)^2 = (h-2)^2 + (k-3)^2$,which simplifies to $x^2 + y^2 - 2hx - 2ky + 4h + 6k - 13 = 0$.This circle cuts $x^2 + y^2 - 12 = 0$ orthogonally.The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here,$g_1 = -h, f_1 = -k, c_1 = 4h + 6k - 13$ and $g_2 = 0, f_2 = 0, c_2 = -12$.Substituting these values: $2(-h)(0) + 2(-k)(0) = (4h + 6k - 13) - 12$.$0 = 4h + 6k - 25$.Replacing $(h, k)$ with $(x, y)$,the locus is $4x + 6y - 25 = 0$.

If a system of circles passes through $(2,3)$ and cuts the circle $x^2+y^2=12$ orthogonally,then the equation of the locus of the centres of that system of circles is:

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