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A line is drawn through a fixed point $P(\alpha, \beta)$ to cut the circle $x^{2}+y^{2}=r^{2}$ at $A$ and $B$. Then
$P A \cdot P B$ is equal to
Options:
$P A \cdot P B$ is equal to
Solution:
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Verified Answer
The correct answer is:
$\alpha^{2}+\beta^{2}-r^{2}$
The equation of any line through the point $P(\alpha, \beta)$ is
$$
\frac{x-\alpha}{\cos \theta}=\frac{y-\beta}{\sin \theta}=k \text { (say) }
$$
Any point on this line is
$$
(\alpha+k \cos \theta, \beta+k \sin \theta)
$$
This point lies on the given circle, if
$$
(\alpha+k \cos \theta)^{2}+(\beta+k \sin \theta)^{2}=r^{2}
$$
or $k^{2}+2 k(\alpha \cos \theta+\beta \sin \theta)$
$$
+\alpha^{2}+\beta^{2}-r^{2}=0
$$
Which being quadratic in $k$, gives two values of $k$. Let $P A=k_{1}, P B=k_{2}$, where $k_{1}, k_{2}$ are the roots of Eq. (i), then
$$
P A \cdot P B=k_{1} k_{2}=\alpha^{2}+\beta^{2}-r^{2}
$$
$$
\frac{x-\alpha}{\cos \theta}=\frac{y-\beta}{\sin \theta}=k \text { (say) }
$$
Any point on this line is
$$
(\alpha+k \cos \theta, \beta+k \sin \theta)
$$
This point lies on the given circle, if
$$
(\alpha+k \cos \theta)^{2}+(\beta+k \sin \theta)^{2}=r^{2}
$$
or $k^{2}+2 k(\alpha \cos \theta+\beta \sin \theta)$
$$
+\alpha^{2}+\beta^{2}-r^{2}=0
$$
Which being quadratic in $k$, gives two values of $k$. Let $P A=k_{1}, P B=k_{2}$, where $k_{1}, k_{2}$ are the roots of Eq. (i), then
$$
P A \cdot P B=k_{1} k_{2}=\alpha^{2}+\beta^{2}-r^{2}
$$
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