For two points $A(2,1)$ and $B(1,2), P$ is a point such that $P A: P B=2: 1$, then locus of $P$ is
For two points $A(2,1)$ and $B(1,2), P$ is a point such that $P A: P B=2: 1$, then locus of $P$ is
$3 x^2+3 y^2+4 x+14 y-15=0$
$3 x^2+3 y^2-4 x-14 y+15=0$
$3 x^2+3 y^2+2 x+7 y+13=0$
$3 x^2+3 y^2-2 x-7 y-13=0$
Solution
Two points $A(2,1)$ and $B(1,2)$ and another point $P$, is such that $P A: P B=2: 1$.
Let us assume $P$ is $(x, y)$
Now, using distance formula
and
$
\begin{aligned}
& P A=\sqrt{(x-2)^2+(y-1)^2} \\
& P(B)=\sqrt{(x-1)^2+(y-2)^2} \\
& \frac{P A}{P B}=\frac{2}{1}=\frac{\sqrt{(x-2)^2+(y-1)^2}}{\sqrt{(x-1)^2+(y-2)^2}}=\frac{2}{1}
\end{aligned}
$
Squaring both sides
$
\begin{aligned}
& \frac{(x-2)^2+(y-1)^2}{(x-1)^2+(y-2)^2}=\frac{4}{1} \\
\Rightarrow & x^2-4 x+4+y^2-2 y+1 \\
& =4\left(x^2-2 x+1+y^2-4 y+4\right) \\
\Rightarrow & x^2-4 x+4+y^2-2 y+1 \\
& =4 x^2-8 x+4+4 y^2-16 y+16 \\
\Rightarrow & 3 x^2+3 y^2-4 x-14 y+15=0
\end{aligned}
$