If (h, k) is the centre of the circle which p | vedclass

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(C) The equation of a circle passing through the origin $(0,0)$ and cutting the circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ or

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-$1$

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(C) The equation of a circle passing through the origin $(0,0)$ and cutting the circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ orthogonally is given by the determinant equation: $\left|\begin{array}{ccc} x^2+y^2 & x & y \ c_1 & g_1 & f_1 \ c_2 & g_2 & f_2 \end{array}ight| = 0$ Here,$c_1=12, g_1=2, f_1=3$ and $c_2=9, g_2=2, f_2=-3$. Substituting these values: $\left|\begin{array}{ccc} x^2+y^2 & x & y \ 12 & 2 & 3 \ 9 & 2 & -3 \end{array}ight| = 0$ Expanding the determinant: $(x^2+y^2)(-6-6) - x(-36-27) + y(24-18) = 0$ $-12(x^2+y^2) + 63x + 6y = 0$ $x^2+y^2 - \frac{63}{12}x - \frac{6}{12}y = 0$ $x^2+y^2 - \frac{21}{4}x - \frac{1}{2}y = 0$ The centre $(h, k)$ is $(\frac{21}{8}, \frac{1}{4})$. Thus,$k-2h = \frac{1}{4} - 2(\frac{21}{8}) = \frac{1}{4} - \frac{21}{4} = -\frac{20}{4} = -5$.

If $(h, k)$ is the centre of the circle which passes through the origin and cuts the circles $x^2+y^2+4x+6y+12=0$ and $x^2+y^2+4x-6y+9=0$ orthogonally,then $k-2h=$

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