A line is drawn through a fixed point P (α , β) to cut the circle x2 + y 2 = r2 at A and B . Then PA ⋅ PB is equal to

Question

A line is drawn through a fixed point P (α , β) to cut the circle x2 + y 2 = r2 at A and B . Then PA ⋅ PB is equal to (A)  (α+β)2 -r2  (B)  α2+β2 -r2  (C)  (α- β)2 +r2  (D) None of the above

Tardigrade Admin · Accepted Answer

The equation of any line through the point

P (α, β)

is

\frac{x - α}{c o s θ} = \frac{y - β}{s i n θ} = k

(say)
Any point on this line is

(α + k cos θ, β + k s in θ)

This point lies on the given circle, if

(α + k cos θ)^{2} + (β + k sin θ)^{2} = r^{2}

or

k^{2} + 2 k (α cos θ + β sin θ)

+ α^{2} + β^{2} - r^{2} = 0 \dots (i)

Which being quadratic in

k

, gives two values of

k

.
Let

P A = k_{1}, PB = K_{2}

, where

k_{1}, K_{2}

2 are the roots of Eq. (i), then

P A \cdot PB = k_{1} k_{2} = α^{2} + β^{2} - r^{2}

Q. A line is drawn through a fixed point $P (α, β)$ to cut the circle $x^{2} + y^{2} = r^{2}$ at $A$ and $B$ . Then $P A \cdot PB$ is equal to

$(α + β)^{2} - r^{2}$

$α^{2} + β^{2} - r^{2}$

$(α - β)^{2} + r^{2}$

None of the above

Q. A line is drawn through a fixed point P(α,β) to cut the circle x2+y2=r2 at A and B . Then PA⋅PB is equal to

(α+β)2−r2

α2+β2−r2

(α−β)2+r2

None of the above

Q. A line is drawn through a fixed point $P (α, β)$ to cut the circle $x^{2} + y^{2} = r^{2}$ at $A$ and $B$ . Then $P A \cdot PB$ is equal to

$(α + β)^{2} - r^{2}$

$α^{2} + β^{2} - r^{2}$

$(α - β)^{2} + r^{2}$