Question

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

Answer 1

Hint:
We will begin by letting \[x = h\] and the corresponding value of $y$ from the equation of line $2x + y = 4$. Then, substitute this point in the equation of the chord of contact, $x{x_1} + y{y_1} = {a^2}$, where $a$ is the radius of the circle. Also, if the equation of the \[P + \lambda Q = 0\], then it passes through the point $P = 0$ and $Q = 0$.

Complete step by step solution:
We are given the equation of line is $2x + y = 4$
Let \[x = h\], then
$
  2h + y = 4 \\
   \Rightarrow y = 4 - 2h \\
$
We know that the equation of chord of contact is $x{x_1} + y{y_1} = {a^2}$, where $a$ is the radius of the circle.
Then, we will substitute the values of \[{x_1} = h\] and ${y_1} = 4 - 2h$ in the equation of chord of contact of the required circle.
$
  x\left( h \right) + y\left( {4 - 2h} \right) = 1 \\
   \Rightarrow \left( {4y - 1} \right) + h\left( {x - 2y} \right) = 0 \\
$
Now, the equation of the \[P + \lambda Q = 0\], then it passes through the point $P = 0$ and $Q = 0$
Then
$
  4y - 1 = 0 \\
   \Rightarrow y = \dfrac{1}{4} \\
$
And
$
  x - 2y = 0 \\
   \Rightarrow x - 2\left( {\dfrac{1}{4}} \right) = 0 \\
   \Rightarrow x = \dfrac{1}{2} \\
$

Hence, the coordinate of point of contact of chord to the circle is $\left( {\dfrac{1}{2},\dfrac{1}{4}} \right)$

Note:
The standard equation of the circle is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, where $r$ is the radius of the circle and $\left( {h,k} \right)$ are the coordinates of centre of circle. Then, the end point of a chord will always lie on the circle and will satisfy the equation of the circle.