Circle 4 Question 2
2. The line $x=y$ touches a circle at the point $(1,1)$. If the circle also passes through the point $(1,-3)$, then its radius is
(2019 Main, 10 April I)
(a) $3 \sqrt{2}$
(b) $2 \sqrt{2}$
(c) 2
(d) 3
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Answer:
Correct Answer: 2. (b)
Solution:
- Since, the equation of a family of circles touching line $L=0$ at their point of $\operatorname{contact}\left(x _1, y _1\right)$ is $\left(x-x _1\right)^{2}+\left(y-y _1\right)^{2}+\lambda L=0$, where $\lambda \in R$.
$\therefore$ Equation of circle, touches the $x=y$ at point $(1,1)$ is
$$ (x-1)^{2}+(y-1)^{2}+\lambda(x-y)=0 $$
$\Rightarrow x^{2}+y^{2}+(\lambda-2) x+(-\lambda-2) y+2=0$
$\because$ Circle (i) passes through point $(1,-3)$.
$$ \begin{array}{rlrl} & & & 1+9+(\lambda-2)+3(\lambda+2)+2=0 \\ \Rightarrow & & 4 \lambda+16 & =0 \\ \Rightarrow & & \lambda=-4 \end{array} $$
So, equation of circle (i) at $\lambda=-4$, is
$$ x^{2}+y^{2}-6 x+2 y+2=0 $$
Now, radius of the circle $=\sqrt{9+1-2}=2 \sqrt{2}$.
