TS EAMCET · Maths · Circle
The equation of the line perpendicular to the radical axis of two circles \(x^2+y^2-5 x+6 y+12=0, x^2+y^2+6 x-\) \(4 y-14=0\) and passing, through \((1,1)\) is
- A \(2 x+3 y-5=0\)
- B \(x+y-2=0\)
- C \(10 x+11 y-21=0\)
- D \(11 x+10 y-21=0\)
Answer & Solution
Correct Answer
(C) \(10 x+11 y-21=0\)
Step-by-step Solution
Detailed explanation
Radical axis of \(S_1\) and \(S_2\) is \(S_1-S_2=0\) i.e. \(\left(x^2+y^2-5 x+6 y+12\right)-\left(x^2+y^2+6 x-4 y-14\right)=0\) \[ \begin{aligned} & \Rightarrow \quad-11 x+10 y+26=0 \\ & \therefore \quad m_1=\frac{11}{10} \end{aligned} \] \(\therefore \quad\) Line perpendicular…
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