Straight Lines
Section Formula — Centre of Locus
nta_pyq_2023_apr
Grade None
Question:
Let $A=(1,2)$ and $B$ any point on $x^2+y^2=16$. If $P$ divides $AB$ in ratio $3:2$ and centre of locus of $P$ is $C(\alpha,\beta)$, then $|AC|$ is equal to
\dfrac{3\sqrt{5}}{5}
\dfrac{4\sqrt{5}}{5}
\dfrac{2\sqrt{5}}{5}
\dfrac{6\sqrt{5}}{5}
Step-by-Step Solution
Key Concept: $P=\frac{2A+3B}{5}$. As $B$ traces circle $|B|=4$: $P-\frac{2A}{5}=\frac{3B}{5}\Rightarrow|P-\frac{2A}{5}|=\frac{12}{5}$. Centre $C=\frac{2A}{5}=(\frac{2}{5},\frac{4}{5})$.
$C=(\frac{2}{5},\frac{4}{5})$. $|AC|=\frac{3\sqrt{5}}{5}$.
Correct Answer: 1