TS EAMCET · Maths · Circle
If a point \(\mathrm{P}\) moves so that the distance from \((0,2)\) to \(\mathrm{P}\) is \(\frac{1}{\sqrt{2}}\) times the distance of \(\mathrm{P}\) from \((-1,0)\), then. the locus of the point \(\mathrm{P}\) is
- A a circle with centre \((1,4)\) and radius 10 units
- B a circle with centre \((-1,-4)\) and radius \(\sqrt{10}\) units
- C a circle with centre \((1,4)\) and radius \(\sqrt{10}\) units
- D a parabola with focus at \((1,4)\) and length of latus rectum 10 units
Answer & Solution
Correct Answer
(C) a circle with centre \((1,4)\) and radius \(\sqrt{10}\) units
Step-by-step Solution
Detailed explanation
\(\mathrm{P}(x, y), \mathrm{A}(0,2), \mathrm{B}(-1,0)\) \[ \begin{aligned} & \mathrm{PA}=\frac{1}{\sqrt{2}} \mathrm{~PB} \\ & \therefore 2 \mathrm{PA}^2=\mathrm{PB}^2 \\ & 2\left[(x-0)^2+(y-2)^2\right]=(x+1)^2+y^2 \\ & x^2+y^2-2 x-8 y+7=0 \end{aligned} \] It is a circle centre…
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