WBJEE · Maths · Circle
A chord \(A B\) is drawn from the point \(A(0,3)\) on
the circle \(x^{2}+4 x+(y-3)^{2}=0,\) and is extended to \(M\) such that \(A M=2 A B\). The locus
of \(M\) is
- A \(x^{2}+y^{2}-8 x-6 y+9=0\)
- B \(x^{2}+y^{2}+8 x+6 y+9=0\)
- C \(x^{2}+y^{2}+8 x-6 y+9=0\)
- D \(x^{2}+y^{2}-8 x+6 y+9=0\)
Answer & Solution
Correct Answer
(C) \(x^{2}+y^{2}+8 x-6 y+9=0\)
Step-by-step Solution
Detailed explanation
Given, \(A M=2 A B\) \(\Rightarrow B\) is mid-point of \(A M\) \(\therefore\) Coordinate of \(B\) is \(\left(\frac{0+x_{1}}{2}, \frac{3+y_{1}}{2}\right)\) \[ =\left(\frac{x_{1}}{2}, \frac{y_{1}+3}{2}\right) \] since, \(B\) lies on the circle \(x^{2}+4 x+(y-3)^{2}=0\)…
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