JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let PQ and MN be two straight lines touching the circle \( x^{2}+y^{2}-4x-6y-3=0 \) at the points A and B respectively. Let O be the centre of the circle and \( \angle AOB=\pi/3. \) Then the locus of the point of intersection of the lines PQ and MN is:
- A \( 3(x^{2}+y^{2})-18x-12y+25=0 \)
- B \( x^{2}+y^{2}-12x-18y-25=0 \)
- C \( x^{2}+y^{2}-18x-12y-25=0 \)
- D \( 3(x^{2}+y^{2})-12x-18y-25=0 \)
Answer & Solution
Correct Answer
(D) \( 3(x^{2}+y^{2})-12x-18y-25=0 \)
Step-by-step Solution
Detailed explanation
Given Circle: \(x^2+y^2-4 x-6 y-3=0\) \(C(2,3) \&\ r=4\) \(\cos 30^{\circ}=\frac{ r }{ OR }=\frac{4}{ OR }\) \(\Rightarrow O R=\frac{8}{\sqrt{3}}\) Now \(OR ^2=( h -2)^2+( k -3)^2\) \(\Rightarrow 3\left(x^2+y^2\right)-12 x-18 y-25=0\)
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