Question

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

Answer 1

Circle : $(x-3{)}^2+y^2=9$
Parabola : $y^2=4x$
Let common tangent equation be $y=mx+\frac{a}{m}$
$\Rightarrow y=mx+\frac{1}{m}$
$\Rightarrow m^{2}x-my+1=0$
the above line is also tangent to circle
$(x-3{)}^2+y^2=9$
So, perpendicular distance from (3, 0) to line $=3$
$\Rightarrow \left|\frac{3m^2-0+1}{\sqrt{m^2+m^4}}\right|=3$
$\Rightarrow (3m^2+1{)}^2=9(m^2+m^4)$
$\Rightarrow m=\pm \frac{1}{\sqrt{3}}$
$\therefore$ tangent equation is
$y=\frac{1}{\sqrt{3}}x+\sqrt{3}$
slope shouldnot be negetive for point of contacts in first quadrent
$\therefore m=\frac{1}{\sqrt{3}}$

For Parabola, point of contact is $(\frac{a}{m^2},\frac{2a}{m})\equiv (3,2\sqrt{3})\equiv (c,d)$

Solving Circle $(x-3{)}^2+y^2=9$ & line equation $y=\frac{1}{\sqrt{3}}x+\sqrt{3}$
$(x-3{)}^2+{(\frac{1}{\sqrt{3}}x+\sqrt{3})}^2=9$
$\Rightarrow x^2+9-6x+\frac{1}{3}{x}^2+3+2x=9$
$\Rightarrow \frac{4}{3}{x}^2-4x+3=0$
$\Rightarrow x=\frac{3}{2}=a$
$\therefore 2(a+c)=2(\frac{3}{2}+3)=9$