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

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(C) The given equations of the circles are:$2x^2 + 2y^2 - 3x + 6y + k = 0$Dividing by $2$,we get:$x^2 + y^2 - \frac{3}{2}x + 3y + \frac{k}{2} = 0$ ...

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$4$

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(C) The given equations of the circles are:$2x^2 + 2y^2 - 3x + 6y + k = 0$Dividing by $2$,we get:$x^2 + y^2 - \frac{3}{2}x + 3y + \frac{k}{2} = 0$ .....$(i)$And the second circle is:$x^2 + y^2 - 4x + 10y + 16 = 0$ .....$(ii)$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 cut orthogonally,the condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Comparing $(i)$ and $(ii)$ with the standard form:$g_1 = -\frac{3}{4}, f_1 = \frac{3}{2}, c_1 = \frac{k}{2}$$g_2 = -2, f_2 = 5, c_2 = 16$Substituting these values into the condition:$2(-\frac{3}{4})(-2) + 2(\frac{3}{2})(5) = \frac{k}{2} + 16$$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 $x^2 + y^2 - 4x + 10y + 16 = 0$ cut orthogonally,then the value of $k$ is

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