Question

Tangents are drawn to a unit circle with centre at the origin from each point on the line ${\rm{2x + y = 4}}$. Find the equation to the locus of the middle point of the chord of contact is

Answer 1

Hint:

Here in this question we have to firstly write the equation of the circle and line. Then we have to find out the equation of the chord of contact with respect to point of contact on the line. Then we again have to find out the equation of the chord of contact with its middle point coordinates and then by taking the ratios of the coefficients of the ${\rm{x and y}}$ as both the equation is of the same line. Then by solving this ratio of the coefficients we will get the equation to the locus of the middle point of the chord of contact.

Complete step by step solution:

It is given that the circle is a unit circle with center at the origin. Therefore, the equation of the circle is

${{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}{\rm{ = 1}}$

Given equation of the line is ${\rm{2x + y = 4}}$

As the point A ${\rm{(}}{{\rm{x}}_1}{\rm{,}}{{\rm{y}}_1}{\rm{)}}$ lies on the line therefore it satisfies the equation of the line. So, we get

${\rm{2}}{{\rm{x}}_1}{\rm{ + }}{{\rm{y}}_1}{\rm{ = 4}}$

$\Rightarrow {{\rm{y}}_1}{\rm{ = 4}} - {\rm{2}}{{\rm{x}}_1}$………. (1)

Now we have to write the equation of the chord of contact PQ with respect to point A. So, we get

${\rm{x}}{{\rm{x}}_1}{\rm{ + y}}{{\rm{y}}_1}{\rm{ = 1}}$………. (2)

Now we have to write the equation of the chord of contact PQ with middle point R ${\rm{(h, k)}}$. So, the equation is tangent equals to ${{\rm{S}}_1}$.

$\Rightarrow {\rm{T = }}{{\rm{S}}_1}$

$\Rightarrow {\rm{xh + yk - 1 = }}{{\rm{h}}^2}{\rm{ + }}{{\rm{k}}^2}{\rm{ - 1}}$

$\Rightarrow {\rm{hx + ky = }}{{\rm{h}}^2}{\rm{ + }}{{\rm{k}}^2}$………. (3)

We know that both equation (2) and equation (3) is the equation of the same line. So we can take the ratios of the coefficients of the ${\rm{x and y}}$, we get

$\Rightarrow \dfrac{{{{\rm{x}}_1}}}{{\rm{h}}}{\rm{ = }}\dfrac{{{{\rm{y}}_1}}}{{\rm{k}}}{\rm{ = }}\dfrac{{\rm{1}}}{{{{\rm{h}}^2}{\rm{ + }}{{\rm{k}}^2}}}$

From equation (1) we will put the value of ${{\rm{y}}_1}$ in the above equation, we get

$\Rightarrow \dfrac{{{{\rm{x}}_1}}}{{\rm{h}}}{\rm{ = }}\dfrac{{{\rm{4}} - {\rm{2}}{{\rm{x}}_1}}}{{\rm{k}}}{\rm{ = }}\dfrac{{\rm{1}}}{{{{\rm{h}}^2}{\rm{ + }}{{\rm{k}}^2}}}$…………(4)

Now we have to solve this equation (4) by equating the first two terms of the equation, we get

$\Rightarrow \dfrac{{{{\rm{x}}_1}}}{{\rm{h}}}{\rm{ = }}\dfrac{{{\rm{4}} - {\rm{2}}{{\rm{x}}_1}}}{{\rm{k}}}$

$\Rightarrow {\rm{k}}{{\rm{x}}_{\rm{1}}}{\rm{ = 4h}} - {\rm{2h \times }}{{\rm{x}}_{\rm{1}}}$

Now by simplifying the above equation, we get

$\Rightarrow ({\rm{k + 2h}}){{\rm{x}}_{\rm{1}}}{\rm{ = 4h}}$

$\Rightarrow {{\rm{x}}_{\rm{1}}}{\rm{ = }}\dfrac{{{\rm{4h}}}}{{({\rm{k + 2h}})}}$

Again we have to solve this equation (4) by equating the first terms and the last term of the equation, we get

$\Rightarrow \dfrac{{{{\rm{x}}_1}}}{{\rm{h}}}{\rm{ = }}\dfrac{{\rm{1}}}{{{{\rm{h}}^2}{\rm{ + }}{{\rm{k}}^2}}}$

Now in this equation we have to put the value of ${{\rm{x}}_1}$ which we calculated earlier, we get

$\Rightarrow \dfrac{{{\rm{4h}}}}{{({\rm{k + 2h}}){\rm{h}}}}{\rm{ = }}\dfrac{{\rm{1}}}{{{{\rm{h}}^2}{\rm{ + }}{{\rm{k}}^2}}}$

$\Rightarrow 4({{\rm{h}}^2}{\rm{ + }}{{\rm{k}}^2}) = {\rm{k + 2h}}$

Now we know that ${\rm{(h, k)}}$ are the x and y coordinates of the middle point of the chord. So we can replace h with x and k with y. Therefore, we get

$\Rightarrow 4({{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}) = {\rm{y + 2x}}$

$\Rightarrow 4({{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}) - {\rm{2x}} - {\rm{y = 0}}$

Hence, $4({{\rm{x}}^2}{\rm{ + }}{{\rm{y}}^2}) - {\rm{2x}} - {\rm{y = 0}}$ is the equation to the locus of the middle point of the chord of contact.

Note:

Chord of a circle is the line segment whose endpoint always lies on the circumference of the circle and it should be noted that the diameter of the circle is the longest chord which passes through the centre of the circle or we can say that the chords 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.