WBJEE · Maths · Circle
A point moves, so that the sum of squares of its distance from the points (1,2) and (-2,1) is always \(6 .\) Then, its locus is
- A the straight line \(y-\frac{3}{2}=-3\left(x+\frac{1}{2}\right)\)
- B a circle with centre \(\left(-\frac{1}{2}, \frac{3}{2}\right)\) and radius \(\frac{1}{\sqrt{2}}\)
- C a parabola with focus (1,2) and directrix passing through (-2,1)
- D an ellipse with foci (1,2) and (-2,1)
Answer & Solution
Correct Answer
(B) a circle with centre \(\left(-\frac{1}{2}, \frac{3}{2}\right)\) and radius \(\frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Let \(P\) be any point, whose coordinate is \((h, k)\) Given. P moves, so that the sum of squares of its distances from the points \(A(1,2)\) and \(B(-2,1)\) is \(6 .\) 1.e., \((P A)^{2}+(P B)^{2}=6\) \(\Rightarrow \quad(h-1)^{2}+(k-2)^{2}+(h+2)^{2}+(k-1)^{2}=6\)…
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