JEE Main 2022 — Mathematics Coordinate Geometry

The circle ${x}^{2}+{y}^{2}+6x+8y+16=0$x2+y2+6x+8y+16=0 has centre $(-3,-4)$(−3,−4) and radius $\sqrt{9+16-16}=3$9+16−16​=3 units.

The circle ${x}^{2}+{y}^{2}+2(3-\sqrt{3})x+2(4-\sqrt{6})y=$x2+y2+2(3−3​)x+2(4−6​)y= $k+6\sqrt{3}+8\sqrt{6},k>0$k+63​+86​,k>0 has centre $(\sqrt{3}-3,\sqrt{6}-4)$(3​−3,6​−4) and radius $\sqrt{{(\sqrt{3}-3)}^{2}+{(\sqrt{6}-4)}^{2}+k+6\sqrt{3}+8\sqrt{6}}=\sqrt{k+34}$(3​−3)2+(6​−4)2+k+63​+86​​=k+34​

Given that these two circles touch internally, so

distance between their centres$=|\text{difference of radii}|$=∣difference of radii∣

$$\sqrt{3+6}=|\sqrt{k+34}-3|$$3+6​=∣k+34​−3∣

$$\Rightarrow \sqrt{k+34}-3=\pm 3$$⇒k+34​−3=±3

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

Now the equation of common tangent to both the circles is given by $2\sqrt{3}x+2\sqrt{6}y+16+k+6\sqrt{3}+8\sqrt{6}=0$23​x+26​y+16+k+63​+86​=0

$\because k=2$∵k=2 then equation becomes

$$x+\sqrt{2}y+3\sqrt{3}+3+4\sqrt{2}=0\cdots (i)$$x+2​y+33​+3+42​=0⋯(i)

$\because (\alpha ,\beta )$∵(α,β) are foot of perpendicular from $(-3,-4)$(−3,−4) to this common tangent, then

$$\frac{\alpha +3}{1}=\frac{\beta +4}{\sqrt{2}}=\frac{-(-3-4\sqrt{2}+3+4\sqrt{2}+3\sqrt{3})}{1+2}$$1α+3​=2​β+4​=1+2−(−3−42​+3+42​+33​)​

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

$\Rightarrow {(\alpha +\sqrt{3})}^{2}=9$⇒(α+3​)2=9 and ${(\beta +\sqrt{6})}^{2}=16$(β+6​)2=16

Hence, ${(\alpha +\sqrt{3})}^{2}+{(\beta +\sqrt{6})}^{2}=25$(α+3​)2+(β+6​)2=25