Q89 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
$x-2 y-6=0$ is a normal to the circle $x^2+y^2+2 g x+2 f y-8=0$. If the line $y=2$ touches this circle, then the radius of the circle can be
- A$\sqrt{32}$
- B$6$
- Correct answerC$4$
- D$\sqrt{18}$
Solution
Centre of circle is $(-g,-f)$
It lies on the normal, $-\mathrm{g}+2 f-6=0$ $\begin{aligned} & \Rightarrow|-f-2|=\sqrt{g^2+f^2-C} \\ & \Rightarrow f^2+4+4 f=g^2+f^2+8 \Rightarrow g^2+4=4 f=2(g+6) \\ & \Rightarrow g^2+4=2 g+12 \Rightarrow g^2-2 g-8=0 \\ & \Rightarrow(g-4)(g+2)=0 \Rightarrow g=4,-2 \end{aligned}$
Case 1: $g=4 \Rightarrow f=5 \Rightarrow r=\sqrt{16+25+8}=7$ Case 2: $g=-2 \Rightarrow f=2 \Rightarrow r=\sqrt{4+4+8}=4$.
It lies on the normal, $-\mathrm{g}+2 f-6=0$ $\begin{aligned} & \Rightarrow|-f-2|=\sqrt{g^2+f^2-C} \\ & \Rightarrow f^2+4+4 f=g^2+f^2+8 \Rightarrow g^2+4=4 f=2(g+6) \\ & \Rightarrow g^2+4=2 g+12 \Rightarrow g^2-2 g-8=0 \\ & \Rightarrow(g-4)(g+2)=0 \Rightarrow g=4,-2 \end{aligned}$
Case 1: $g=4 \Rightarrow f=5 \Rightarrow r=\sqrt{16+25+8}=7$ Case 2: $g=-2 \Rightarrow f=2 \Rightarrow r=\sqrt{4+4+8}=4$.
Hence, the correct answer is option: 'C'.