JEE Main 2022 — Circle Question with Solution

The circle $$x^2+y^2+6x+8y+16=0$$ has centre $$(-3,-4)$$ and radius 3 units.

The circle $$x^2+y^2+2\left(3-\sqrt{3}\right)x+2\left(4-\sqrt{6}\right)y=k+6\sqrt{3}+8\sqrt{6},k>0$$ has centre $$\left(\sqrt{3}-3,\sqrt{6}-4\right)$$ and radius $$\sqrt{k+34}$$

$$\because$$ These two circles touch internally hence

$$\sqrt{3+6}=\left|\sqrt{k+34}-3\right|$$

Here, $$k=2$$ is only possible ($$\because$$ $$k>0$$)

Equation of common tangent to two circles is $$2\sqrt{3}x+2\sqrt{6}y+16+6\sqrt{3}+8\sqrt{6}+k=0$$

$$\because$$ $$k=2$$ then equation is

$$x+\sqrt{2}y+3+4\sqrt{2}+3\sqrt{3}=0$$ ...... (i)

$$\because$$ ($$\alpha$$, $$\beta$$) are foot of perpendicular from $$(-3,-4)$$

To line (i) then

$$\frac{\alpha +3}{1}=\frac{\beta +4}{\sqrt{2}}=\frac{-\left(-3-4\sqrt{2}+3+4\sqrt{2}+3\sqrt{3}\right)}{1+2}$$

$$\therefore$$ $$\alpha +3=\frac{\beta +4}{\sqrt{2}}=-\sqrt{3}$$

$$\Rightarrow {\left(\alpha +\sqrt{3}\right)}^2=9$$ and $${\left(\beta +\sqrt{6}\right)}^2=16$$

$$\therefore$$ $${\left(\alpha +\sqrt{3}\right)}^2+{\left(\beta +\sqrt{6}\right)}^2=25$$