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.

• Have you noticed the figure forms right triangles? – N.S.JOHN Nov 26 '16 at 16:10
• That's a terribly uninformative title. – user137731 Nov 26 '16 at 16:32
• I'm sorry, I didn't want to make it seem like I was just begging for the right answer. – cgug123 Nov 26 '16 at 16:36
• @cgug123: You could still do that by having a flatly descriptive title such as "Distance between two tangent points on a circle" or something like that. – Brian Tung Nov 26 '16 at 17:12
• The problem has one trivial solution unless we can assume that the points $P$ and $Q$ are distinct. – Sid Nov 26 '16 at 17:34

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 $PA=\sqrt 3$

• I can see that it is, and that the angle is 60 degrees, but I don't know how to get the side lengths from there. Sorry, I'm very inexperienced in geometrical applications – cgug123 Nov 26 '16 at 16:18
• @cgug123 Got the equilateral triangle? – Qwerty Nov 26 '16 at 16:18
• Yes. I'm not sure how to prove it is equilateral like the first answerer, however. I can just see that it is. – cgug123 Nov 26 '16 at 16:21
• @cgug123 Whats $\angle BPQ$? Can youprove it is same as $\angle PAB$? – Qwerty Nov 26 '16 at 16:21
• 60 degrees, yes? – cgug123 Nov 26 '16 at 16:22

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 $PO_2$.

• Okay this I can do! Thank you. – cgug123 Nov 26 '16 at 16:35

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.