TS EAMCET · Maths · Circle
The number of common tangents of the circles \(x^2+y^2-4\) \(=0\) and \(x^2+y^2-6 x-8 y-24=0\) is
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
\(x^2+y^2=4\) \(\begin{aligned} & C_1:(0,0) \\ & r_1=2 \\ & x^2+y^2-6 x-8 y-24=0 \\ & C_2:(3,4) \\ & r_2=7 \\ & C_1 C_2=\sqrt{(3-0)^2+(4-0)^2}= \\ & \Rightarrow C_1 C_2=\left(r_1-r_2\right)\end{aligned}\) \(\therefore \quad\) They are touching each other internally.…
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