JEE Main 2023 — Circle Question with Solution

Given,

$A$ be the point $\left(1,2\right)$ and $B$ be any point on the curve $x^2+y^2=16$,

And the centre of the locus of the point $P$, which divides the line segment $AB$ in the ratio $3:2$ is the point $C\left(\alpha ,\beta \right)$,

Now let the point on the circle $x^2+y^2=16$ be $B\left(4\cos \theta ,4\sin \theta \right)$

Now using section formula in $A\left(1,2\right)$ and $B\left(4\cos \theta ,4\sin \theta \right)$ we get,

$P\left(\frac{12\cos \theta +2}{5},\frac{12\sin \theta +4}{5}\right)\equiv (h,k)$

$\Rightarrow \cos \theta =\frac{5h-2}{12} \& \sin \theta =\frac{5k-4}{12}$

Now squaring and adding we get,

${\left(\frac{5h-2}{12}\right)}^2+{\left(\frac{5k-4}{12}\right)}^2=1$

$\Rightarrow {\left(h-\frac{2}{5}\right)}^2+{\left(k-\frac{4}{5}\right)}^2={\left(\frac{12}{5}\right)}^2$

So, the centre of the locus is $C\left(\frac{2}{5},\frac{4}{5}\right)$

Hence, by distance formula we get,$AC=\sqrt{{\left(\frac{3}{5}\right)}^2+{\left(\frac{6}{5}\right)}^2}=\frac{3\sqrt{5}}{5}$