TS EAMCET · Maths · Circle
For all real values of \(k\), the polar of the point (2k,k-4) with respect to \(x^2+y^2-4 x-6 y+1=0\) passes through the point
- A \((1,1)\)
- B \((1,-1)\)
- C \((-3,1)\)
- D \((3,1)\)
Answer & Solution
Correct Answer
(D) \((3,1)\)
Step-by-step Solution
Detailed explanation
The polar of point \(P(2 k, k-4)\) with respect to the circle, \[ \begin{aligned} & x^2+y^2-4 x-6 y+1=0 \text { is } \\ & \begin{aligned} & 2 k x+y(k-4)+(-2)(x+2 k) \\ &+(-3)\{y+(k-4)\}+1=0 \\ & \Rightarrow k(2 x+y-7)-7 y-2 x+13=0 \end{aligned} \end{aligned} \] Hence, \((3,1)\)…
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