Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1400 Upvotes
Verified Answer
The correct answer is:
a circle with centre $(1,4)$ and radius $\sqrt{10}$ units
$\mathrm{P}(x, y), \mathrm{A}(0,2), \mathrm{B}(-1,0)$
$$
\begin{aligned}
& \mathrm{PA}=\frac{1}{\sqrt{2}} \mathrm{~PB} \\
& \therefore 2 \mathrm{PA}^2=\mathrm{PB}^2 \\
& 2\left[(x-0)^2+(y-2)^2\right]=(x+1)^2+y^2 \\
& x^2+y^2-2 x-8 y+7=0
\end{aligned}
$$
It is a circle
centre $(1,4)$
$$
\text { radius }=\sqrt{1^2+4^2-7}=\sqrt{10}
$$
$$
\begin{aligned}
& \mathrm{PA}=\frac{1}{\sqrt{2}} \mathrm{~PB} \\
& \therefore 2 \mathrm{PA}^2=\mathrm{PB}^2 \\
& 2\left[(x-0)^2+(y-2)^2\right]=(x+1)^2+y^2 \\
& x^2+y^2-2 x-8 y+7=0
\end{aligned}
$$
It is a circle
centre $(1,4)$
$$
\text { radius }=\sqrt{1^2+4^2-7}=\sqrt{10}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.