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Let $Q$ be a point on the circle $B: x^2+y^2=a^2$ and $P(h, k)$ be a fixed point. If the locus of the point which divides the join of $P$ and $Q$ in the ratio $p: q$ is a circle $C$, then the centre of $C$ is
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Verified Answer
The correct answer is:
$\left(\frac{h q}{p+q}, \frac{k q}{p+q}\right)$
Let $R(\alpha, \beta)$ be the required point and $Q\left(x_0, y_0\right)$ and $P(h, k)$.
$$
\begin{aligned}
& \therefore \quad(\alpha, \beta) & =\left(\frac{p x_0+q h}{p+q}, \frac{p y_0+q k}{p+q}\right) \\
\Rightarrow & \alpha & =\frac{p x_0+q h}{p+q} \text { and } \beta=\frac{p y_0+q k}{p+q} \\
\Rightarrow & x_0 & =\frac{(p+q) \alpha-q h}{p} \text { and } y_0=\frac{(p+q) \beta-q k}{p}
\end{aligned}
$$
Since, $Q\left(x_0, y_0\right)$ lies on the circle $x^2+y^2=a^2$
$$
\begin{aligned}
& \therefore\left(\frac{(p+q) \alpha-q h}{p}\right)^2+\left(\frac{(p+q) \beta-q k}{p}\right)^2=a^2 \\
& \Rightarrow\left(\alpha-\frac{q h}{p+q}\right)^2+\left(\beta-\frac{q k}{p+q}\right)^2=\frac{p^2 a^2}{(p+q)^2}
\end{aligned}
$$
$\therefore$ Locus of $R(\alpha, \beta)$ is
$$
\left(x-\frac{q h}{p+q}\right)^2+\left(y-\frac{q k}{p+q}\right)^2=\frac{p^2 a^2}{(p+q)^2}
$$
which is a circle whose centre is $\left(\frac{q h}{p+q}, \frac{q k}{p+q}\right)$.
$$
\begin{aligned}
& \therefore \quad(\alpha, \beta) & =\left(\frac{p x_0+q h}{p+q}, \frac{p y_0+q k}{p+q}\right) \\
\Rightarrow & \alpha & =\frac{p x_0+q h}{p+q} \text { and } \beta=\frac{p y_0+q k}{p+q} \\
\Rightarrow & x_0 & =\frac{(p+q) \alpha-q h}{p} \text { and } y_0=\frac{(p+q) \beta-q k}{p}
\end{aligned}
$$
Since, $Q\left(x_0, y_0\right)$ lies on the circle $x^2+y^2=a^2$
$$
\begin{aligned}
& \therefore\left(\frac{(p+q) \alpha-q h}{p}\right)^2+\left(\frac{(p+q) \beta-q k}{p}\right)^2=a^2 \\
& \Rightarrow\left(\alpha-\frac{q h}{p+q}\right)^2+\left(\beta-\frac{q k}{p+q}\right)^2=\frac{p^2 a^2}{(p+q)^2}
\end{aligned}
$$
$\therefore$ Locus of $R(\alpha, \beta)$ is
$$
\left(x-\frac{q h}{p+q}\right)^2+\left(y-\frac{q k}{p+q}\right)^2=\frac{p^2 a^2}{(p+q)^2}
$$
which is a circle whose centre is $\left(\frac{q h}{p+q}, \frac{q k}{p+q}\right)$.
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