I have a circle of undetermined size. If I draw a line straight down from the top of the circle, and then draw a line at a 90 degree angle, I will have defined an arc where the lines intersect the circle. If I know the length of the two lines (say 9 units down, 32 units across) can I find the radius of the circle? If so, how?
If you are saying that the first line starts on the circle, and you go 9 down and 32 across, and have a second point where the line hits the circle, you have only defined two points on the circle and there are an infinite number of circles that include those two points. You need three points on the circle to define it, unless you have additional information.
You could also have the center and one point on the circle, or you could have two points that are opposite one another, and define the circle with two points that way.
It can be done. Suppose that your circle’s centre is in (0,0). The top of the circle is therefore (0,r), where r is the radius. Call dy the length of your line going down, and dx the length of your line going left or right. You know that both (-dx, r-dy) and (dx, r-dy) are points on the circle. Using the formula for the circle, you should be able to isolate r.
He does have three points on the circle, since he’s starting from the top of the circle. That’s a point, there’s the point at 9 down and 32 right from there, and a point 9 down and 32 left as well.
If the first line is indeed a diameter in the circle (I suppose “straight down from the top” is fairly unambiguous) you can use pythagoras.
If you’re asking what you seem to be asking, you can do it with high school trig. Let a be the length of the line segment along the diameter of the circle descending from the point at the top, b is the length of the line segment at a right angle extending to the edge of the circle, and imagine the hypotenuse formed by drawing a line between the two points on your circle. We will call this triangle A. Imagine another radius drawn from the center of the circle out to the corner of A where your right angle segment intersects the circle. The radius your original line segment lies along, and the new radius form an isoceles triangle whose base angle is the corner of A near the top of the circle. That hypotenuse of A, length sqrt(a^2+b^2), is the base of that isoceles triangle, and the base angle is atan(b/a). You now have enough to solve for the radius as the length of a congruent side of an isoceles triangle with base sqrt(a^2 + b^2), and base angle atan(b/a).
Much easier to draw than describe.
I agree.
A relatively easy solution involves a seldom-used geometric theorem which states that if two chords of a circle intersect, the product of the line segments are equal.
Of course, I have a diagram: 
The chords are 64 and the other “chord” is actually the diameter.
If we multiply the segments, we get the equation:
32 * 32 = 9 * (Diameter - 9)
1,024 = 9 * Diameter -81
1,105 = 9 * d
diameter = 122.777
and therefore the radius = 61.389
A couple of thoughts. You can also say the radius is the solution to
r^2 = 32^2 + (r - 9)^2
by the pythagorean theorem.
However, the way I was taught was to find something that was a (very large) rectangle and lay it over the circle so a corner just touched the circle. Then the other two points where the rectangle exited the circle would lie on the ends of a diameter.
pmwgreen
Good solution.
I figured I’d make another diagram:
A carpenter’s square is ideal for this, and is one of the standard tricks it is used for. Do it twice, and you can find the center.