QJEEAdv12Ma8782: JEE Advanced 2012 Mathematics - Class 11 JEEAdv12Ma8782 Solution: A tangent PT is drawn to the circle at the point P . A straight …

The equation of tangent of the circle $x^2+y^2=4$ is $y=m x \pm 2 \sqrt{1+m^2}\quad \text{.... (i)}$

Let $y=m x \pm 2 \sqrt{1+m^2}$ also touches $(x-3)^2+ y^2=1$

$\Rightarrow(x-3)^2+\left(m x \pm 2 \sqrt{1+m^2}\right)^2=1$

$\Rightarrow x^2-6 x+9+m^2 x^2+4\left(1+m^2\right) \pm 4 m \sqrt{1+m^2} x=1$

$\Rightarrow\left(1+m^2\right) x^2+\left(-6 \pm 4 m \sqrt{1+m^2}\right) x+4\left(m^2+3\right)=0$

Apply

$\begin{aligned}& \quad \left(-6 \pm 4 m \sqrt{1+m^2}\right)^2-4\left(1+m^2\right) \cdot 4\left(m^2+3\right)=0 \\\Rightarrow \quad & 36+16 m^2\left(1+m^2\right) \pm 48 m \sqrt{1+m^2} \\& \quad -16\left(m^4+4 m^2+3\right)=0 \\\Rightarrow \quad & 4 m^2+1= \pm 4 m \sqrt{1+m^2}\end{aligned}$

On squaring both side

$\begin{array}{rlrl}\Rightarrow & 16 m^4+1+8 m^2 =16 m^2+16 m^4 \\\Rightarrow & m^2 =\frac{1}{8} \\\Rightarrow & m = \pm \frac{1}{2 \sqrt{2}}\end{array}$

$\begin{array}{ll}\text { Put } & m= \pm \frac{1}{2 \sqrt{2}} \text { in the equation (i) } \\\Rightarrow & y= \pm \frac{x}{2 \sqrt{2}} \pm \frac{6}{2 \sqrt{2}} \\\Rightarrow & 2 \sqrt{2} y= \pm x \pm 6\end{array}$

Hence, the equation of common tangent of given circles are $2 \sqrt{2} y=-x+6,2 \sqrt{2} y=x+6, 2 \sqrt{2} y=-x-6$ and $2 \sqrt{2} y=x-6$ .