Question

The equation of the common tangent to $y^2=8x$ and $x^2+y^2-12x+4=0$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Equation of a tangent to $y^2=8x$ is $y=mx+\frac{2}{m}$
Perpendicular distance of a tangent from centre of the circle is equal to its radius.
This touches $x^2+y^2-12x+4=0\Rightarrow \sqrt{36-4}=\frac{|6m+\frac{2}{m}|}{\sqrt{1+m^2}}\Rightarrow m^4-2m^2+1=0\Rightarrow (m^2-1{)}^2=0\Rightarrow m^2=1\Rightarrow m=\pm 1$
The common tangents are $y=\pm (x+2)$