TS EAMCET · Maths · Circle
The radical centre of the circles \(x^2+y^2-4 x-6 y+5=0\), \(x^2+y^2-2 x-4 y-1=0\) and \(x^2+y^2-6 x-2 y=0=0\) lies on the line
- A \(x+y-5=0\)
- B \(2 x-4 y+7=0\)
- C \(4 x-6 y+5=0\)
- D \(18 x-12 y+1=0\)
Answer & Solution
Correct Answer
(D) \(18 x-12 y+1=0\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & S_1=x^2+y^2-4 x-6 y+5=0 \\ & S_2=x^2+y^2-2 x-4 y-1=0 \\ & S_3=x^2+y^2-6 x-2 y=0 \\ & S_1-S_2=0 \Rightarrow 2 x+2 y-6=0 \Rightarrow x+y-3=0 \\ & S_2-S_3=0 \Rightarrow 4 x-2 y-1=0 \end{aligned} \] Solving Eqs. (i) and (ii), we get…
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