TS EAMCET · Maths · Circle
A point \(P\) moves so that distance from \((0,2)\) to \(P\) is \(\frac{1}{\sqrt{2}}\) times the distance of \(P\) from \((-1,0)\). Then the locus of the point is
- A a circle with centre at \((1,4)\) and radius \(\sqrt{10}\)
- B a parabola with focus at \((1,4)\) and length of latus rectum 10
- C an ellipse with centre at \((-1,-4)\) and length of the major axis \(\sqrt{10}\)
- D a hyperbola with centre at \((-1,-4)\) and length of the transverse axis 10
Answer & Solution
Correct Answer
(A) a circle with centre at \((1,4)\) and radius \(\sqrt{10}\)
Step-by-step Solution
Detailed explanation
Let point \(P\) is \((h, k)\). Given \(Q(0,2)\) and \(R(-1,0)\) then given \(P Q=\frac{1}{\sqrt{2}} P R\) Squaring on both sides, \(2(P Q)^2=(P R)^2\) \(2\left(h^2+(k-2)^2\right)=(h+1)^2+k^2\) \(2 h^2+2 k^2-8 k+8=h^2+1+2 h+k^2\) \(\therefore \quad h^2+k^2-2 h-8 k+7=0\)…
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