For what value of $k$ the circles $x^2+y^2+5 x+3 y+7=0$ and $x^2+y^2-8 x+6 y+k=0$ cuts orthogonally

Question

Aicharya · Accepted Answer

Let the two circles be $x^2+y^2+2 g_1 x+2 f_1 y$ $+c_1=0$ and $x^2+y^2+2 g_2 x+2 f_2 y+c_2=0$ where $g_1=5 / 2, f_1=3 / 2, c_1=7, g_2=4, f_2=3$ and $\mathrm{c}_2=\mathrm{k}$ If the two circles intersects orthogonally, then $\begin{aligned} & 2~\left(\mathrm{g}_1 \mathrm{~g}_2+\mathrm{f}_1 \mathrm{f}_2\right)=\mathrm{c}_1+\mathrm{c}_2 \Rightarrow 2\left(-10+\frac{9}{2}\right)=7+\mathrm{k} \ & \Rightarrow 11=7+\mathrm{k} \Rightarrow \mathrm{k}=-18 \end{aligned}$

For what value of \(k\) the circles \(x^2+y^2+5 x+3 y+7=0\) and \(x^2+y^2-8 x+6 y+k=0\) cuts orthogonally

Solution