Question

Find locus of a point from which chord of contact to circle $x^2+y^2=4$ touches the circle $(x-2{)}^2+y^2=4$

Answer 1

Let $P(a,b)$ be the point from which the chord of contact to the circle $x^2+y^2=4$ i.e., $x^2+y^2=2^2$ …………$\left(1\right)$

Touches the circle $(x-2{)}^2+y^2=4$ i.e., $(x-2{)}^2+y^2=2^2$ ………..$\left(2\right)$

Equation of the chord of contact to circle $\left(1\right)$ wrt P, we have $ax+by=4$

$\Rightarrow y=\frac{4-ax}{b}$

Since the chord of contact touches the circle $\left(2\right)$, we can write $\left(2\right)$ as.

$(x-2{)}^2+{\left(\frac{4-ax}{b}\right)}^2=4$

$\Rightarrow x^2+4+4x+\frac{16+a^2{x}^2-8ax}{b^2}=4$

$\Rightarrow b^2{x}^2+4b^2-4b^{2}x+16+a^2{x}^2-8ax=4b^2$

$\Rightarrow (a^2+b^2)x^2-(4b^2+8a)x=16=0$

Since the chord of contact touches the circle $\left(2\right)$ at a single point.

$\therefore$ The above quadratic equation must have equal roots.

$\therefore \{-(4b^2+8a){\}}^2=\mathrm{4.16.}(a^2+b^2)=0$

$\Rightarrow 4^2{b}^4+64a^2+64a{b}^2-64a^2-64b^2=0$

$\Rightarrow 16b^4+64a{b}^2-64b^2=0$

$\Rightarrow 16(b^4+4a{b}^2-4b^2)=0$

$\Rightarrow b^4+4a{b}^2-4b^2=0$

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

$\Rightarrow b^2+4a-4=0$ …………..$\left(3\right)$

Substituting $a=x$ and $b=y$ in equation $\left(3\right)$, we get $y^2+4x-4=0$, which is the required locus.