JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 1)
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:
4
6
2
1
Explanation
$$\begin{gathered} C_1: x^2+y^2-2(x+y)+1=0 \\ C_2:(x+1)^2+y^2=(2)^2 \\ x^2+y^2+2 x-3=0 \end{gathered}$$
Common chord is
$$\begin{aligned} & C_1-C_2=0 \\ & \Rightarrow 2 x+y-2=0 \end{aligned}$$
also, this line intersects the $$y$$-axis at the point
$$\begin{aligned} & P(y, 0) . \\ & \Rightarrow y=2 \end{aligned}$$
$$P(2,0)$$
Distance of point $$P$$ from $$(1,1)$$ is
$$\begin{aligned} & d=\sqrt{(2-1)^2+(0-1)^2} \\ & =\sqrt{1^2+1^2} \\ & d=\sqrt{2} \\ & \Rightarrow d^2=2 \\ \end{aligned}$$
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