Question

The chords of contact of the pair of tangents drawn from each point on the line $2x + y = 4$ to the circle ${x^2} + {y^2} = 1$ pass through the point.

Answer 1

Parametrize the point on the line and find the chords of contact in that specific point. For finding the chords of contact we can go with $T = 0$ to find our desired result.

Complete step by step solution:
Let, $(h,k)$ be any point on the line $2x + y = 4$.
So, we can say, $2h + k = 4$
$\Rightarrow k = 4 - 2h$
So, we get a point on the line $2x + y = 4$ is of the form $(h,4 - 2h)$ .
Equation of the chord of contact on the point $(h,k)$ ,
$hx + ky = 1$
$\Rightarrow hx + (4 - 2h)y = 1$ as $k = 4 - 2h$
$\Rightarrow hx + 4y - 2hy - 1 = 0$
$\Rightarrow (4y - 1) + h(x - 2y) = 0$ …….. (1)
We have it as of the form, $P + \lambda Q = 0$ .
So, we pass through the intersection of $P = 0$ and $Q = 0$
Now if we see equation (1), we can fix that,
This line passes through the point of intersection of $(4y - 1) = 0$ and $(x - 2y) = 0$
As,

$$4y - 1 = 0 \\\ \Rightarrow 4y = 1 \\\ \Rightarrow y = \dfrac{1}{4} \\\$$

And then,

$$x - 2y = 0 \\\ \Rightarrow x - 2 \times \dfrac{1}{4} = 0 \\\ \Rightarrow x - \dfrac{1}{2} = 0 \\\ \Rightarrow x = \dfrac{1}{2} \\\$$

That is, through the point, $\left( {\dfrac{1}{2},\dfrac{1}{4}} \right)$ .

The chords of contact of the pair of tangents drawn from each point on the line $2x + y = 4$ to the circle ${x^2} + {y^2} = 1$ pass through the point, $\left( {\dfrac{1}{2},\dfrac{1}{4}} \right)$

Note: We can find the chord of contact also by putting T = 0 easily. Chord of contact is always associated with the transverse axis. Here we have parameterized the problem to deal with it and find the solution.