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AP EAMCET · Maths · Circle

The equation of a tangent to the circle \(x^2+y^2+2 x-12 y\) \(-132=0\) which is perpendicular to the line \(12 x+5 y+k=0\) is

  1. A \(5 x-12 y+92=0\)
  2. B \(5 \mathrm{x}-12 \mathrm{y}-246=0\)
  3. C \(5 \mathrm{x}-12 \mathrm{y}-169=0\)
  4. D \(5 x-12 y+246=0\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(5 x-12 y+246=0\)

Step-by-step Solution

Detailed explanation

Given equation of circle \(\begin{aligned} & x^2+y^2+2 x-12 y-132=0 \\ & \Rightarrow(x+1)^2+(y-6)^2=13^2 \\ & \Rightarrow \text { Radius }=13, \text { Centre }=(-1,6) \end{aligned}\) Since slope of the line \(12 x+5 y+8=0\) is \(\mathrm{m}_1=\frac{-12}{5}\) So slope of…