If the circles $x^2+y^2+2 \alpha x+2 y-8=0$ and $x^2+y^2-2 x$ $+\alpha y-14=0$ intersect orthogonally, then…
If the circles $x^2+y^2+2 \alpha x+2 y-8=0$ and $x^2+y^2-2 x$ $+\alpha y-14=0$ intersect orthogonally, then the distance between their centres is
- $\sqrt{242}$
- $\sqrt{970}$
- $\sqrt{629}$
- $\sqrt{541}$
Solution
$S_1 \equiv x^2+y^2+2 \alpha x+2 y-8=0$
$\begin{aligned}
& \Rightarrow \quad\left(g_1, f_1, c_1\right)=(\alpha, 1,-8) \\
& S_2 \equiv x^2+y^2-2 x+\alpha y-14=0 \\
& \Rightarrow \quad\left(g_2, f_2, c_2\right)=\left(-1, \frac{\alpha}{2},-14\right)
\end{aligned}$
$S_1$ and $S_2$ cuts orthogonally $\Rightarrow 2 g_1 g_2+2 f_1 f_2=c_1+c_2$
$\Rightarrow-2 \alpha+\alpha=-8-14 \Rightarrow \alpha=22$
Centre of $S_1\left(C_1\right)=(-22,-1)$
Centre of $S_2\left(C_2\right)=(1,-11)$
$C_1 C_2=\sqrt{(23)^2+(10)^2}=\sqrt{629} \text {. }$
Asked in: AP EAMCET 2024 (23 May Shift 1)
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