MHT CET · Maths · Circle
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
- A \((\alpha+\beta)^{2}-r^{2}\)
- B \(\alpha^{2}+\beta^{2}-r^{2}\)
- C \((\alpha-\beta)^{2}+r^{2}\)
- D None of the above
Answer & Solution
Correct Answer
(B) \(\alpha^{2}+\beta^{2}-r^{2}\)
Step-by-step Solution
Detailed explanation
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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