let $y^2=4ax$ be a parabola and $x^2+y^2 +2bx=0$ be a circle. parabola and circle touch each other externally

Question

the parabola and circle touch each other externally, the focus of parabola and centre of circle lie on $x$ axis, then $a$ and $b$ must be both greater than $0$ or both must be less than $0$ (if they have same signs then only they can touch externally because centre of circle is $(-b,0)$.

but when i solve the two curves and equate discriminant to $0$ then signs of $a$ and $b$ come out be opposite . why is that?

Answer 1

HINT:

Any point on $y^2=x$ is $(at^2,2at)$

So, the equation of the tangent $$y(2at)=4a\cdot\dfrac{x+at^2}2\iff x-yt+at^2=0$$

It will also be a tangent of $$(x+b)^2+y^2=b^2$$

if $$|b|=\dfrac{-b+0(-t)+at^2}{\sqrt{1+t^2}}$$

$$\iff b^2(1+t^2)=(at^2-b)^2$$

If $t\ne0,$ $$\iff\dfrac{b(b+2a)}{a^2}=t^2>0$$

Answer 2

Consider the equations $$y^2=4ax\tag 1$$ $$x^2+y^2 +2bx=0\tag 2$$ Replace $y^2$ from $(1)$ into $(2)$. This gives $$x^2+(4a+2b)x=0 \tag 3$$ the solution of which being $$x_1=0 \qquad, \qquad x_2=-2(2a+b)$$ Using thesese values in $(1)$, we then have $$y_1=0 \qquad, \qquad y_{2,3}=\pm 2 \sqrt{2} \sqrt{-a (2 a+b)}$$