The radical centre of the circles $x^2+y^2+2 x+3 y+1=0$, $x^2+y^2+x-y+3=0, x^2+y^2-3 x+2 y+5=0$ is
The radical centre of the circles $x^2+y^2+2 x+3 y+1=0$, $x^2+y^2+x-y+3=0, x^2+y^2-3 x+2 y+5=0$ is
- $\left(-\frac{7}{38}, \frac{6}{19}\right)$
- $\left(\frac{6}{19}, \frac{14}{19}\right)$
- $\left(\frac{14}{19}, \frac{6}{19}\right)$
- $\left(\frac{2}{19}, \frac{3}{19}\right)$
Solution
$\mathrm{S}_1 \equiv x^2+y^2+2 x+3 y+1=0$
$\begin{aligned} & \mathrm{S}_2 \equiv x^2+y^2+x-y+3=0 \\ & \mathrm{~S}_3 \equiv x^2+y^2-3 x+2 y+5=0\end{aligned}$
$\begin{aligned}
& \mathrm{S}_1-\mathrm{S}_2=0 \Rightarrow x+4 y-2=0 \quad \ldots (i) \\
& \mathrm{~S}_2-\mathrm{S}_3=0 \Rightarrow 4 x-3 y-2=0 \quad \ldots (ii)
\end{aligned}$
Solving (i) and (ii) we get
$\text { radical centre }=\left(\frac{14}{19}, \frac{6}{19}\right)$
Asked in: AP EAMCET 2024 (20 May Shift 2)
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