Question
The chords of contact of the pair of tangents to the circle $${x^2} + {y^2} = 1$$ drawn from any point on the line $$2x + y = 4$$ pass through the point :
A.
$$\left( {\frac{1}{2},\,\frac{1}{4}} \right)$$
B.
$$\left( {\frac{1}{4},\,\frac{1}{2}} \right)$$
C.
$$\left( {1,\,\frac{1}{2}} \right)$$
D.
$$\left( {\frac{1}{2},\,1} \right)$$
Answer :
$$\left( {\frac{1}{2},\,\frac{1}{4}} \right)$$
Solution :
Any point on $$2x + y = 4$$ is $$\left( {t,\,4 - 2t} \right).$$ The equation of the chord of contact of the point is $$x.t + y.\left( {4 - 2t} \right) = 1{\text{ or }}4y - 1 + t\left( {x - 2y} \right) = 0$$
This line passes through the intersection of $$4y - 1 = 0$$ and $$x - 2y = 0.$$