TS EAMCET · Maths · Circle
\(L_1\) and \(L_2\) are two common tangents to two circles. If \(L_1\) touches the two circles at \(A(1,1)\) and \(B(0,1)\) and \(L_2\) touches the two circles at \(C\left(\frac{3}{5}, \frac{4}{5}\right), D\left(\frac{-1}{5}, \frac{7}{5}\right)\), then the equation of the radical axis of the two circles is
- A \(2 x-6 y=7\)
- B \(2 x+y+7=0\)
- C \(2 x+6 y=7\)
- D \(x=y\)
Answer & Solution
Correct Answer
(C) \(2 x+6 y=7\)
Step-by-step Solution
Detailed explanation
As radical axis of the two circles bisects the all common tangents of the circles. Now, the mid-point of \(A(1,1)\) and \(B(0,1)\) is \(M\left(\frac{1}{2}, 1\right)\) and mid-point of \(C\left(\frac{3}{5}, \frac{4}{5}\right)\) and \(D\left(-\frac{1}{5}, \frac{7}{5}\right)\) is…
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