TS EAMCET · Maths · Circle
The point/points of intersection of the common tangents of the two circles \(x^2+y^2-8 x-6 y+21=0\) and \(x^2+y^2-2 y-15=0\) is/are
- A \((5,8),(-4,3)\)
- B \((8,5)\)
- C \((3,1)\)
- D \((2,1),(4,3)\)
Answer & Solution
Correct Answer
(B) \((8,5)\)
Step-by-step Solution
Detailed explanation
Given circle, \(S_1: x^2+y^2-8 x-6 y+21=0\) and \(S_2: x^2+y^2-2 y-15=0\) Circle \(S_1\) : Centre \(C_1(4,3)\), radius \(r_1=\sqrt{16+9-21}=2\) \(S_2:\) Centre \(C_2(0,1)\), radius \(r_2=\sqrt{1+15}=4\) Now \(\quad C_1 C_2=\sqrt{4^2+2^2}=\sqrt{20}\)…
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