Question

For the two circle $x^2+y^2=16$ and $x^2+y^2-2y=0$, there is/ are:-

• A. no common tangent
• B. two pairs of common tangents
• C. one pair of common tangents
• D. three common tangents

Answer 1

The correct option is A no common tangent

$x^2+y^2-2y=0$

$\Rightarrow x^2+(y-1{)}^2=1$

Equation of tangent to the circle :

$\Rightarrow y-1=mx\pm 1\sqrt{1+m^2}$

$\Rightarrow y=mx\pm \sqrt{1+m^2}+1$

Equation of tangent to $x^2+y^2=16$

$y=mx\pm 4\sqrt{1+m^2}$

If both the tangents have to be same , then

$\pm \sqrt{1+m^2}+1=\pm 4\sqrt{1+m^2}$

$\Rightarrow \pm 3\sqrt{1+m^2}=1$

$\Rightarrow \pm \sqrt{1+m^2}=\frac{1}{3}$

$\Rightarrow \sqrt{1+m^2}=\pm \frac{1}{3}$

i.e.$\sqrt{1+m^2}=\frac{1}{3}$

and $\sqrt{1+m^2}=-\frac{1}{3}$

$\therefore \sqrt{1+m^2}$ cant ne negative.

$\therefore$ No common tangent is possible.