The radical centre of the circles $x^2+y^2+3 x+2 y+1=0$ $x^2+y^2-x+6 y+5=0$ and $x^2+y^2+5 x-8 y+15=0$ is
The radical centre of the circles $x^2+y^2+3 x+2 y+1=0$ $x^2+y^2-x+6 y+5=0$ and $x^2+y^2+5 x-8 y+15=0$ is
- $(3,2)$
- $(-3,-2)$.
- $(2,3)$
- $(-2,-3)$
Solution
$
\begin{aligned}
& \text { Let } S_1: x^2+y^2+3 x+2 y+1=0 \\
& S_2: x^2+y^2-x+6 y+5=0 \\
& S_3: x^2+y^2+5 x-8 y+15=0
\end{aligned}
$
Let $(x, y)$ be the radical centre, then $S_1-S_2=0$ and
$
\begin{aligned}
& S_2-S_3=0 \\
& \Rightarrow \quad 4 x-4 y-4=0 \text { and }-6 x+14 y-10=0 \\
& \Rightarrow \quad x=y+1 \text { and } 3 x-7 y+5=0 \\
& \Rightarrow \quad 3(y+1)-7 y+5=0 \Rightarrow 3 y+3-7 y+5=0 \\
& \Rightarrow \quad-4 y+8=0 \\
& \therefore \quad y=2 \text { and } x=2+1=3
\end{aligned}
$
Radical centre is $(3,2)$
Asked in: AP EAMCET 2022 (06 Jul Shift 2)
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