Q90 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
The slope of a common tangent to the circles $x^2+y^2-4 x-8 y+16=0$ and $x^2+y^2-6 x-16 y+64=0$ is
- A0
- Correct answerB15/8
- C1
- D17/4
Solution
$\begin{aligned}
& \text { (b) } x^2+y^2-4 x-8 y+16=0 \Rightarrow C_1 \equiv(2,4), n=42 \\
& x^2+y^2-6 x-16 y+64=0 \Rightarrow C_1 \equiv(3,8), r_2=3
\end{aligned}$
Let $m x-y+C=0$ be the common tangent to the two circles. $\begin{aligned} & \left|\frac{2 m-4+C}{\sqrt{1+m^2}}\right|=2 \Rightarrow C=2 \sqrt{1+m^2}-2 m+4 \\ & \left|\frac{3 m-8+C}{\sqrt{1+m^2}}\right|=3 \Rightarrow C=3 \sqrt{1+m^2}-3 m+8 \\ & 2 \sqrt{1+m^2}-2 m+4=3 \sqrt{1+m^2}-3 m+8 \\ & \Rightarrow \sqrt{1+m^2}=m-4 \Rightarrow 1+m^2=m^2-8 m+16 \\ & \Rightarrow 8 m=15 \Rightarrow m=\frac{15}{8} \end{aligned}$
Let $m x-y+C=0$ be the common tangent to the two circles. $\begin{aligned} & \left|\frac{2 m-4+C}{\sqrt{1+m^2}}\right|=2 \Rightarrow C=2 \sqrt{1+m^2}-2 m+4 \\ & \left|\frac{3 m-8+C}{\sqrt{1+m^2}}\right|=3 \Rightarrow C=3 \sqrt{1+m^2}-3 m+8 \\ & 2 \sqrt{1+m^2}-2 m+4=3 \sqrt{1+m^2}-3 m+8 \\ & \Rightarrow \sqrt{1+m^2}=m-4 \Rightarrow 1+m^2=m^2-8 m+16 \\ & \Rightarrow 8 m=15 \Rightarrow m=\frac{15}{8} \end{aligned}$
Hence, the correct answer is option: 'B'.