Question

The chords of contact of the pair of tangents drawn from each point on the line \[2x + y = 4\] to the circle \[{x^2} + {y^2} = 1\] passes through a fixed point \[M\left( {{x_0},{y_0}} \right)\]. The value of \[\left( {{x_0} + 6{y_0}} \right)\] is equal to:
A.0
B.1
C.2
D.3

Answer 1

Hint: Here we will firstly find the value of \[x\] and \[y\] from the equation of line. Then we will write the standard equation of the chord of contact and then put the value of \[x\] and \[y\] in it. Then we will solve the equation to get two equations and by solving that we will get the value of the coordinates of the point M. Then we will put the value of \[{x_0},{y_0}\] in the equation \[\left( {{x_0} + 6{y_0}} \right)\] to get the required value.

Complete step-by-step answer:
The given equation of circle is \[{x^2} + {y^2} = 1\] and according to the equation of the circle, we get center of circle as \[\left( {0,0} \right)\] and radius of the circle as 1.
Now we will write the given equation of the line is \[2x + y = 4\].
Let the value of \[x\] in the equation of line be \[a\] i.e. \[x = a\].
Therefore, the value of \[y\] is equal to \[y = 4 - 2x = 4 - 2a\]
Now we will write the equation of the chord of contact and put the value of \[x\] and \[y\] in it. Therefore, we get
Equation of chord of contact \[x{x_1} + y{y_1} = {r^2}\] where, \[r\] is the radius of the circle. Therefore, we get
\[x \times a + y\left( {4 - 2a} \right) = {1^2}\]
Multiplying the terms, we get
\[ \Rightarrow ax + 4y - 2ay - 1 = 0\]
As the chord is passing through the point \[M\left( {{x_0},{y_0}} \right)\]. So, it will satisfy the equation of the chord. Therefore, we get
\[ \Rightarrow a{x_0} + 4{y_0} - 2a{y_0} - 1 = 0\]
Taking out the common terms, we get
\[ \Rightarrow \left( {4{y_0} - 1} \right) + a\left( {{x_0} - 2{y_0}} \right) = 0\]
Therefore, by the equation we will get two equations as
\[4{y_0} - 1 = 0\] and \[{x_0} - 2{y_0} = 0\]
Now by solving the above equation we will get the values of the coordinates of the point \[M\left( {{x_0},{y_0}} \right)\]. Therefore, we get
\[ \Rightarrow {y_0} = \dfrac{1}{4}\] and \[{x_0} = 2{y_0}\]
By putting the value of \[{y_0}\] in the equation \[{x_0} = 2{y_0}\], we will get the value of \[{x_0}\]. Therefore,
\[ \Rightarrow {x_0} = 2 \times \dfrac{1}{4} = \dfrac{1}{2}\]
Hence the point becomes \[M\left( {\dfrac{1}{2},\dfrac{1}{4}} \right)\].
Now we will find the value of \[\left( {{x_0} + 6{y_0}} \right)\].
Substituting \[{x_0} = \dfrac{1}{2}\] and \[{y_0} = \dfrac{1}{4}\] in \[\left( {{x_0} + 6{y_0}} \right)\], we get
\[\left( {{x_0} + 6{y_0}} \right) = \dfrac{1}{2} + 6 \times \dfrac{1}{4}\]
Multiplying the terms, we get
\[ \Rightarrow \left( {{x_0} + 6{y_0}} \right) = \dfrac{1}{2} + \dfrac{3}{2} = 2\]
Hence the value of \[\left( {{x_0} + 6{y_0}} \right)\] is equal to 2.
So, option C is the correct option.

Note: Chord of a circle is the line segment whose endpoint always lies on the circumference of the circle. We should note that the diameter of the circle is the longest chord which passes through the centre of the circle. In other words, we can say that the chord passing through the center of the circle is always the longest chord of that circle. Also radius of the circle is not known as the chord as one end point lies on the center of the circle. We should know that when a point lies on a curve or a line then that point satisfies the equation of that curve or line.