Question

If the chord of contact of tangents from a point $P(x_1,y_1)$ to the circle ${(x-a)}^2+y^2=a^2$ touches the circle ${(x-a)}^2+y^2=a^2$, then find the locus of $(x_1,y_1).$

Answer 1

Given, equation of the circle is given as

$(x-a{)}^2+y^2=a^2$ ………..$\left(1\right)$

$\Rightarrow x^2+a^2-2a+y^2=a^2$

$\Rightarrow x^2+y^2=+2a$

$\Rightarrow x^2+y^2=(\sqrt{2a}{)}^2$

$\therefore$ Equation of the chord of contact of the circle $\left(1\right)$ wrt to $P(x_1,y_1)$ is given by

$x_{1}x+y_{1}y=(\sqrt{2a}{)}^2$

$\Rightarrow x_{1}x+y_{1}y=2a$

$\Rightarrow x_{1}x+y_{1}y-2a=0$ ……….$\left(2\right)$

Since the line $\left(2\right)$ touches the circle $(x-a{)}^2+y^2=a^2$, we get

$a=\frac{|x_1\cdot a+y_1\cdot 0+(-2a)|}{\sqrt{x_1^2+y_1^2}}$

$a=\frac{|a{x}_1-2a|}{\sqrt{x_1^2+y_1^2}}$

Squaring both sides we get

$a^2=\frac{(a{x}_1-2a{)}^2}{(x_1^2+y_1^2)}$

$\Rightarrow a^2(x_1^2+y_1^2)=(a{x}_1-2a{)}^2$

$\Rightarrow a^2{x}_1^2+a^2{y}_1^2=a^2{x}_1^2+4a^2-4a^2{x}_1$

$\Rightarrow a^2{y}_1^2+4a^2{x}_1-4a^2=0$

$\Rightarrow a^2(y_1^2+4x_1-4)=0$

$\Rightarrow y_1^{2}4x_1-4=0$

$\therefore$ Locus of $P(x_1,y_1)$ is given by, $y^2+4x-4=0$

[Selling $x=x_1,y=y_1$].