Question

A line is drawn through a fix 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^(2)+beta^(2)`
B. `alpha^(2)+beta^(2)-a^(2)`
C. `a^(2)`
D. `alpha^(2)+beta^(2)+a^(2)`

Answer 1

Correct Answer - C
The equation of any line through `P(alpha, beta)` is
`(x-alpha)/(cos theta)=(y - beta)/(sin theta)`
The coordinates of any point on this line at a distance r from P are `(alpha+r cos theta, theta, beta+ r sin theta)`. If it lies on the given circle, then
`(alpha + r cos theta )^(2) + (beta + r sin theta)^(2)=a^(2)`
`rArr r^(2)+2r(alpha cos theta + beta sin theta) + alpha^(2)+beta^(2)-a^(2)=0`
This equation gives two values of r, say `r_(1), r_(2)` These two values are lengths PA and PB.
`:. PA xxPB=r_(1)r_(2)=alpha^(2)+beta^(2)-a^(2)`