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 $PA.PB$ 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: (D)

None of these

When a line through a fixed point P cuts the circle in two different points A and B then $PA.PB$ is called power of the point which is equal to $S_1$

hence

$PA.PB=x_1^2+y_1^2-r^2$

replacing by $\alpha ,\beta$

$P=PA.PB=S_1={\alpha }^2+{\beta }^2-r^2=P{T}^2$