Distance between two points on a circle Two circles with radius 1 are tangent to one another. One line passes through the centre of the first circle and is tangent to the second circle at the point $P$. A second line passes through the centre of the first circle and is tangent to the second circle at the point $Q$. Find the distance between $P$ and $Q$. 
This question appeared in a first year calculus exam, and I can't see how I would even use my knowledge in differential calculus to try and solve this. It seems more of a geometry problem, and when I try to draw a diagram I am left at a loss because there's hardly any information given to try and solve. If someone could give me a hint as to how to begin, that'd be great. Thank you. I also wasn't too sure how to tag it, so my apologies. 
 A: Hint. Make a drawing, and by considering its angles, show that the triangle $\triangle O_1PQ$ is equilateral  where $O_1$ is the centre of the first circle. 
Now note that the $\triangle O_1O_2P$ is a right triangle where $O_2$ is the centre of the second circle. Then by using Pythagoras theorem we can find $O_1P$ ($=PQ$) from $O_1O_2$ and $PO_2$.
A: Let the first circle be centred $A$ and second one $B$.

Check that $\angle PAB=\angle QAB=30^\circ$ (See what the lengths of $PA$ and $PB$ are!)
Also check that $PAQ$ forms an equilateral triangle

Edit:
$PA=\sqrt 3, PB=1,AB=2 $ . Let $PQ$ cut $AB$ at $D$. Triangle  $PAB, BPD, QAB,  QBD$ are similar.
$\angle PBD=\angle QBD=60^\circ$

$\angle APD=90-\angle BPD=90^\circ-30^\circ=60$
$\angle AQD=90-\angle BQD=90^\circ-30^\circ=60$

So in triangle $APQ$ all angles are $60^\circ$ and $PA=\sqrt 3$
A: If this is on a calculus exam, then it’s likely that you’re meant to compute the slope of the tangents to the circle via differentiation. Center one of the circles on the origin and place the center of the other circle at $C=(2,0)$.  
Implicitly differentiate $x^2+y^2=1$ to obtain $dy/dx=-x/y$. The slope of the line through $P=(x,y)$ and $C$ is $(y-0)/(x-2)$. Setting these equal to each other yields the equation $$2x-x^2=y^2.$$ Combine this with the equation of the circle and solve the resulting system.
