AP EAMCET · Maths · Circle
Let the locus of the point of intersection of the perpendicular tangents drawn to the circle \(x^2+y^2+6 x\) \(-4 y-12=0\) be the circle \(S\). Then the equation of the tangent drawn to \(\mathrm{S}\) which is perpendicular to the line \(6 \mathrm{x}-4 \mathrm{y}+\mathrm{k}=0\) is
- A \(4 \mathrm{x}+6 \mathrm{y} \pm \sqrt{26}=0\)
- B \(2 x+3 y \pm \sqrt{26}=0\)
- C \(2 x+3 y \pm 5 \sqrt{26}=0\)
- D \(4 x+6 y \pm 5 \sqrt{26}=0\)
Answer & Solution
Correct Answer
(C) \(2 x+3 y \pm 5 \sqrt{26}=0\)
Step-by-step Solution
Detailed explanation
Given circle is \(x^2+y^2+6 x-4 y-12=0\) Centre \(=(-3,2)\); radius \(=5\) \(\therefore\) Centre of circle \((\mathrm{S})=(-3,2)\) Since \(\angle A P B=90^{\circ}\), therefore \(O A P B\) is square and \(O A=5\) \(\therefore\) Radius of circle \((S)=O P=5 \sqrt{2}\) Now,…
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