Let A be the point (1,2) and B be any point on the curve x^2+y^2=16.

Question

Let $A$ be the point $(1,2)$ and $B$ be any point on the curve $x^2+y^2=16$. If the centre of the locus of the point $P$, which divides the line segment $AB$ in the ratio $3:2$ is the point $C(\alpha,\beta)$, then the length of the line segment $AC$ is:

Accepted Answer

We have, equation of circle is $x^2+y^2=16$

Let any point on the circle $x^2+y^2=4^2$ is $B(4 \cos \theta, 4 \sin \theta)$ and $A(1,2)$

Let $\mathrm{P}$ be $(h, k)$ which divides $\mathrm{AB}$ in $3: 2$

So

$$
h=\frac{12 \cos \theta+2}{3+2} \text { and } k=\frac{12 \sin \theta+2 \times 2}{3+2}
$$

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

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

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

$\therefore$ Locus of point $P$ is $\left(x-\frac{2}{5}\right)^2+\left(y-\frac{4}{5}\right)^2=\frac{12^2}{5^2}$

which is equation of circle with centre $\left(\frac{2}{5}, \frac{4}{5}\right)$

Hence, $A C=\sqrt{\left(1-\frac{2}{5}\right)^2+\left(2-\frac{4}{5}\right)^2}=\frac{\sqrt{45}}{5}=\frac{3 \sqrt{5}}{5}$

Solutions · 1