MHT CET · Maths · Circle
The angle between a pair of tangents drawn from a point 'P' to the circle \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0\) is
\(2 \alpha\). The equation of the locus of the point \({ }^{'} P\) ' is
- A \(x^{2}+y^{2}+4 x-6 y+4=0\)
- B \(x^{2}+y^{2}+4 x-6 y-9=0\)
- C \(x^{2}+y^{2}+4 x-6 y-4=0\)
- D \(x^{2}+y^{2}+4 x-6 y+9=0\)
Answer & Solution
Correct Answer
(D) \(x^{2}+y^{2}+4 x-6 y+9=0\)
Step-by-step Solution
Detailed explanation
The equation of a given circle is
\(
x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0
\)
\(\therefore\) Centre \(A(-2,3)\) in \(\Delta A B P\),
\(
\sin \alpha=\frac{2 \sin \alpha}{\sqrt{(h+2)^{2}+(k-3)^{2}}}
\)
\(\Rightarrow (h+2)^{2}+(k-3)^{2}=4\)
\(\Rightarrow h^{2}+k^{2}+4 h-6 k+9=0\)
Hence, the required locus of \(P(h, k)\) is
\(
x^{2}+y^{2}+4 x-6 y+9=0
\)
\(
x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0
\)
\(\therefore\) Centre \(A(-2,3)\) in \(\Delta A B P\),
\(
\sin \alpha=\frac{2 \sin \alpha}{\sqrt{(h+2)^{2}+(k-3)^{2}}}
\)
\(\Rightarrow (h+2)^{2}+(k-3)^{2}=4\)
\(\Rightarrow h^{2}+k^{2}+4 h-6 k+9=0\)
Hence, the required locus of \(P(h, k)\) is
\(
x^{2}+y^{2}+4 x-6 y+9=0
\)
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