JEE Main 2023 (Online) 10th April Evening ShiftQuestion 1
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
Circle
Solution
Concept. Write $P$ by the section formula; as $B$ runs over a circle, $P$ runs over a scaled circle. Formulas used. (1) Point dividing $AB$ in $m:n$: $P=\dfrac{nA+mB}{m+n}$Step 1 - Use (1). $3:2$: $P=\frac{2A+3B}{5}=\frac25A+\frac35B$. Step 2 - Locus. $|B|=4$, so $\big|P-\frac25A\big|=\frac35\cdot4=\frac{12}{5}$: a circle with centre $C=\frac25A=\big(\frac25,\frac45\big)$. Step 3 - $AC$. $A-C=\frac35A$, so $AC=\frac35|A|=\frac35\sqrt5=\frac{3\sqrt5}{5}$. Answer: A