If the two circles 2x^2 + 2y^2 - 3x + 6y + k | vedclass

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(C) The given circles are $2x^2 + 2y^2 - 3x + 6y + k = 0$ and $x^2 + y^2 - 4x + 10y + 16 = 0$.Dividing the first equation by $2$,we get $x^2 + y^2 - \

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$4$

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(C) The given circles are $2x^2 + 2y^2 - 3x + 6y + k = 0$ and $x^2 + y^2 - 4x + 10y + 16 = 0$.Dividing the first equation by $2$,we get $x^2 + y^2 - \frac{3}{2}x + 3y + \frac{k}{2} = 0 \dots (i)$.The second equation is $x^2 + y^2 - 4x + 10y + 16 = 0 \dots (ii)$.For two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to intersect orthogonally,the condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here,$g_1 = -\frac{3}{4}$,$f_1 = \frac{3}{2}$,$c_1 = \frac{k}{2}$ and $g_2 = -2$,$f_2 = 5$,$c_2 = 16$.Substituting these values into the condition:$2(-\frac{3}{4})(-2) + 2(\frac{3}{2})(5) = \frac{k}{2} + 16$.$3 + 15 = \frac{k}{2} + 16$.$18 = \frac{k}{2} + 16$.$\frac{k}{2} = 2$.$k = 4$.

If the two circles $2x^2 + 2y^2 - 3x + 6y + k = 0$ and $x^2 + y^2 - 4x + 10y + 16 = 0$ intersect orthogonally,then the value of $k$ is:

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