JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 11)
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 length of the line segment $$\mathrm{AC}$$ is :
$$\frac{3 \sqrt{5}}{5}$$
$$\frac{6 \sqrt{5}}{5}$$
$$\frac{2 \sqrt{5}}{5}$$
$$\frac{4 \sqrt{5}}{5}$$
Explanation
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 $3: 2$
_10th_April_Evening_Shift_en_11_1.png)
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, $A C=\sqrt{\left(1-\frac{2}{5}\right)^2+\left(2-\frac{4}{5}\right)^2}=\frac{\sqrt{45}}{5}=\frac{3 \sqrt{5}}{5}$
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, $A C=\sqrt{\left(1-\frac{2}{5}\right)^2+\left(2-\frac{4}{5}\right)^2}=\frac{\sqrt{45}}{5}=\frac{3 \sqrt{5}}{5}$
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