JEE Main 2021 (Online) 25th February Evening ShiftQuestion 1
A line is a common tangent to the circle $(x-3)^{2}+y^{2}=9$ and the parabola $y^{2}=4x$. If the two points of contact $(a,b)$ and $(c,d)$ are distinct and lie in the first quadrant, then $2(a+c)$ is equal to $\_\_\_\_\_\_$ .
Circle
Answer$9$
Solution
Formulas used. (1) Tangent to $y^2=4x$ with slope $m$: $y=mx+\frac1m$, touching at $\big(\frac1{m^2},\frac2m\big)$(2) A line touches a circle $\iff$ distance of the centre from it $=$ radius.(3) Point of contact with a circle $=$ foot of perpendicular from the centre to the tangent.Step 1 - Tangent to the parabola. By (1): $mx-y+\frac1m=0$. Step 2 - It also touches the circle. Circle: centre $(3,0)$, radius $3$. By (2): $$\frac{|3m+\frac1m|}{\sqrt{1+m^2}}=3\Rightarrow 9m^2+6+\frac1{m^2}=9+9m^2$$ $$\Rightarrow \frac1{m^2}=3\Rightarrow m^2=\frac13$$ Step 3 - Choose the sign. The parabola contact $\big(\frac1{m^2},\frac2m\big)$ must be in the first quadrant, so $m>0$: $m=\frac1{\sqrt3}$. Contact $(c,d)=(3,\,2\sqrt3)$. Tangent: $y=\frac{x}{\sqrt3}+\sqrt3$, i.e. $x-\sqrt3y+3=0$.Step 4 - Contact with the circle. By (3), foot from $(3,0)$ to $x-\sqrt3y+3=0$: $$(a,b)=(3,0)-\frac{3-0+3}{1+3}\,(1,-\sqrt3)=\Big(\frac32,\frac{3\sqrt3}{2}\Big)$$ It lies in the first quadrant ✓. Step 5 - Required value. $$2(a+c)=2\Big(\frac32+3\Big)=9$$ Answer: 9