The parabola $y^2=4ax$ and circle $x^2+y^2+2bx=0$ have more then one common tangents;,
Then which one is/are right,
$(a)\; ab>0\;\;\;\; (b)\; ab<0\;\;\;\; (c)\; ab<-2\;\;\;\; (d)\; ab>2$
$\bf{My\; Try::}$ Let $y=mx+c$ is a common tangent to the parabola $y^2=4ax\;$
Then Condition of tangency is $$y=mx+\frac{a}{m}\Rightarrow m^2x-my+a=0$$
Which is also tangent to the circle $(x-0)^2+(y-(-b))^2=b^2$
So using Condition of tangency $$\left|\frac{mb+a}{\sqrt{m^4+m^2}}\right| = b\Rightarrow m^2b^2+a^2+2abm = m^4b^2+m^2b^2$$
So we get $$m^4b^2-2abm-a^2=0$$
Now how can i solve after that, Help me, Thanks
You have a typo in your circle equation.
Your Condition of tangency $$\left|\frac{mb+a}{\sqrt{m^4+m^2}}\right| = b\Rightarrow m^2b^2+a^2+2abm = m^4b^2+m^2b^2$$
Should be $$\left|\frac{-m^2b+a}{\sqrt{m^4+m^2}}\right| = b\Rightarrow m^2=\frac{a^2}{2ab+b^2}$$
Does this help?