If 2 x^{2}+2 y^{2}+4 x+5 y+1=0 and 3 x^{2}+3 | vedclass

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(D) The given equations of the circles are:$2 x^{2}+2 y^{2}+4 x+5 y+1=0 \Rightarrow x^{2}+y^{2}+2 x+\frac{5}{2} y+\frac{1}{2}=0 \quad \dots (i)$$3 x^{

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$-\frac{17}{12}$

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(D) The given equations of the circles are:$2 x^{2}+2 y^{2}+4 x+5 y+1=0 \Rightarrow x^{2}+y^{2}+2 x+\frac{5}{2} y+\frac{1}{2}=0 \quad \dots (i)$$3 x^{2}+3 y^{2}+6 x-7 y+3 k=0 \Rightarrow x^{2}+y^{2}+2 x-\frac{7}{3} y+k=0 \quad \dots (ii)$Comparing these with the standard form $x^{2}+y^{2}+2 g x+2 f y+c=0$,we get:For $(i): g_{1}=1, f_{1}=\frac{5}{4}, c_{1}=\frac{1}{2}$For $(ii): g_{2}=1, f_{2}=-\frac{7}{6}, c_{2}=k$The condition for two circles to be orthogonal is $2 g_{1} g_{2}+2 f_{1} f_{2}=c_{1}+c_{2}$.Substituting the values:$2(1)(1)+2\left(\frac{5}{4}\right)\left(-\frac{7}{6}\right)=\frac{1}{2}+k$$2-\frac{35}{12}=\frac{1}{2}+k$$\frac{24-35}{12}=\frac{1}{2}+k$$-\frac{11}{12}=\frac{1}{2}+k$$k=-\frac{11}{12}-\frac{6}{12}=-\frac{17}{12}$

If $2 x^{2}+2 y^{2}+4 x+5 y+1=0$ and $3 x^{2}+3 y^{2}+6 x-7 y+3 k=0$ are orthogonal,then the value of $k$ is

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