Let the circle $C_1: x^2+y^2-2(x+y)+1=0$ and $\mathrm{C_2}$ be a circle having centre at $(-1,0)$ and radius 2 . If the line of the common chord of $\mathrm{C}_1$ and $\mathrm{C}_2$ intersects the $\mathrm{y}$-axis at the point $\mathrm{P}$, then the square of the distance of P from the centre of $\mathrm{C_1}$ is:
Solution
<p>$$\begin{gathered}
C_1: x^2+y^2-2(x+y)+1=0 \\
C_2:(x+1)^2+y^2=(2)^2 \\
x^2+y^2+2 x-3=0
\end{gathered}$$</p>
<p>Common chord is</p>
<p>$$\begin{aligned}
& C_1-C_2=0 \\
& \Rightarrow 2 x+y-2=0
\end{aligned}$$</p>
<p>also, this line intersects the $y$-axis at the point</p>
<p>$$\begin{aligned}
& P(y, 0) . \\
& \Rightarrow y=2
\end{aligned}$$</p>
<p>$P(2,0)$</p>
<p>Distance of point $P$ from $(1,1)$ is</p>
<p>$$\begin{aligned}
& d=\sqrt{(2-1)^2+(0-1)^2} \\
& =\sqrt{1^2+1^2} \\
& d=\sqrt{2} \\
& \Rightarrow d^2=2 \\
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Circles · Topic: Equation of a Circle
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