So I have a question that goes as follows:-
The circle $x^2+y^2+4\lambda x=0$ $(\lambda \in R)$ touches the parabola $y^2=8x$ . The value of $\lambda$ is given by:
The answer is $\lambda \in (0,\infty)$
But how do I visualize this? I can actually think of two cases.
Firstly, the center of the circle is $(-2\lambda,0)$, which is on the x-axis. If this touches the parabola at one point then clearly $$-2\lambda<0$$
$$\implies \lambda >0$$ This happens when the circle is touching the parabola at the vertex.
But, in my mind I can also think of the case where the centre of the circle lies on the positive x axis and touches the parabola in two points. Is this also not a probable case? If it is, what are the bound for $\lambda $ is such a case?
Basically, we want to solve: $$\left\{\begin{matrix} x^2+y^2+4\lambda x=0 \\ y^2=8x \end{matrix}\right.$$ We can substitute $y^2=8x$ and then we have: $$x(x+8+4\lambda)=0$$ This has two solutions: $$x=0\rightarrow y=0$$ Or: $$x=-8-4\lambda \rightarrow y=\sqrt{8(-8-4\lambda)}$$ In particular, in this case, we must have: $$-8-4\lambda>0 \leftrightarrow \lambda<-2$$ If $\lambda\geq-2$, then circle and parabola intersect at $O(0,0)$. If $\lambda<-2$, then the point of intersection is: $$I(-8-4\lambda,\pm\sqrt{8(-8-4\lambda)})$$