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 $$\mathrm{AB}$$ in the ratio $3: 2$ is the point C$$(\alpha, \beta)$$, then the len...

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 $B(4\cos 	heta ,4\sin 	heta )$ and $A(1,2)$

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

So

$h=\frac{12\cos 	heta +2}{3+2}k=\frac{12\sin 	heta +2	imes 2}{3+2}$

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

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

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

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

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

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

Solutions · 1