Question

The equation of common tangent to the circle $x^2+y^2=2$ and parabola $y^2=8x$ is

• A. $y=x+1$
• B. $y=x+2$
• C. $y=x-2$
• D. $y=-x+2$

Answer 1

The correct option is B $y=x+2$

Given, $y^2=8x$

$\therefore 4a=8\Rightarrow a=2$

Any tangent of parabola is,

$y=mx+\frac{a}{m}$

$\Rightarrow mx-y+\frac{2}{m}=0$

If it is a tangent to the circle $x^2+y^2=2$, then perpendicular from centre $(0,0)$ is equal to radius $\sqrt{2}$.

$\therefore \frac{\frac{2}{m}}{\sqrt{m^2+1}}=\sqrt{2}$

$\Rightarrow \frac{4}{m^2}=2(m^2+1)$

$\Rightarrow m^4+m^2-2=0$

$\Rightarrow (m^2+2)(m^2-1)=0$

$\Rightarrow m=\pm 1$

Hence, the common tangents are $y=\pm (x+2)$.