TS EAMCET · Maths · Circle
The line \(3 x-y+k=0\) touches the circle \(x^2+y^2+4 x-6 y+3=0\). If \(k_1, k_2\left(k_1 < k_2\right)\) are the two values of \(k\), then the equation of the chord of contact of the point \(\left(k_1, k_2\right)\) with respect to the given circle is
- A \(19 x+y-18=0\)
- B \(x+19 y-3=0\)
- C \(x+16 y-56=0\)
- D \(20 x+18 y-7=0\)
Answer & Solution
Correct Answer
(C) \(x+16 y-56=0\)
Step-by-step Solution
Detailed explanation
Line \(3 x-y+k=0\) touches the circle \( \begin{aligned} & x^2+y^2+4 x-6 y+3=0 \text { having } \\ & \text { Centre }(-2,3), r=\sqrt{4+9-3}=\sqrt{10} \\ & \because \quad r=\left|\frac{-6-3+k}{\sqrt{10}}\right| \end{aligned} \)…
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