Question

A circle has its center at the origin, and (5, -12) is a point on the circle. How long is the radius of the circle?

Answer 1

Does the pythagorean theorem mean anything to you? Might it be potentially of use here?

Answer 2

Is that way you solve it? I having a hard time understanding the course..

Answer 3

Yeah, just see if you can draw the picture and see how a circle's radius is just the hypotenuse of a right triangle.

Answer 4

so 13?

Answer 5

Yep, perfect. =

Answer 6

Let's first begin by interpretating the problem, recognizing it and of course, recalling what we know in order to find a solution.
Now we are given a circumference and a point belonging to it, so we can use ANALYTIC GEOMETRY to solve it.
Now, it has center O, meaningthat the coordinates are O(0,0) and the blonging point P(5,-12).
The circle is defined by all the point that have the same distance to a point we call center (O, in this case), the distance all points have to thatcenter is called "radius".
Now, I'll make a little proof for you, just because I just had breakfast and I'll leave soon:
Let P1(x1,y1) and P2(x2,y2) arbitrary in the plane:

Now by pythagorean theorem, we say:
\[d(P1,P2)^{2}=(x _{2}-x _{1})^{2}+(y _{2}-y _{1})\]
\[d(P1,P2)=\sqrt{(x _{2}-x _{1})^{2}+(y _{2}-y _{1})^{2}}\]
and that's what we'll call the "distance between two points in the plane" formula.

applying that little formula to the problem, replacing P1 as O(0,0) and P2 as P(5,-12):
\[d(O,P)=\sqrt{(5-0)^{2}+(-12-0)^{2}}\]
\[d(O,P=r\]
PD: "r" states for "radius"

so we can say:
\[r=\sqrt{(5)^{2}+(-12)^{2}}\]
\[r=\sqrt{25+144}\]
\[r=\sqrt{169}\]
\[r=13\]

and with that, we have found the radius we were asked for in the exercise.