TS EAMCET · Maths · Circle
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
- A 0
- B 15/8
- C 1
- D 17/4
Answer & Solution
Correct Answer
(B) 15/8
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { (b) } x^2+y^2-4 x-8 y+16=0 \Rightarrow C_1 \equiv(2,4), n=42 \\ & x^2+y^2-6 x-16 y+64=0 \Rightarrow C_1 \equiv(3,8), r_2=3 \end{aligned}\) Let \(m x-y+C=0\) be the common tangent to the two circles.…
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