Question

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 $P A . P B$ 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 these

Answer 1

Answer: (B)

$\alpha ^ { 2 } + \beta ^ { 2 } - r ^ { 2 }$

Answer 2

Let the equation of line through the point $( \alpha , \beta )$ be $\frac { x - \alpha } { \cos \theta } = \frac { y - \beta } { \sin \theta } = k$ (say) ….(i)

where k is the distance of any point (x, y) on the line from the point $P ( \alpha , \beta )$ . Let this line meets the circle $x ^ { 2 } + y ^ { 2 } = r ^ { 2 }$ at $( \alpha + k \cos \theta , \beta + k \sin \theta )$ .

$\therefore$ $( \alpha + k \cos \theta ) ^ { 2 } + ( \beta + k \sin \theta ) ^ { 2 } = r ^ { 2 }$

or $k ^ { 2 } + 2 ( \alpha \cos \theta + \beta \sin \theta ) k + \left( \alpha ^ { 2 } + \beta ^ { 2 } - r ^ { 2 } \right) = 0$,

which is a quadratic in k. If $k _ { 1 }$ and $k _ { 2 }$ are its roots and the line (i) meets circle at A and B, then $P A = k _ { 1 }$ and $P B = k _ { 2 }$.

$\therefore$ $P A \cdot P B = k _ { 1 } k _ { 2 } =$Products of roots$= \alpha ^ { 2 } + \beta ^ { 2 } - r ^ { 2 }$.

Trick : As we know from figure, .