Question

A circle has a radius of $15.$ Let $\overline{AB}$ be a chord of the circle, such that $AB = 40$. What is the distance between the chord and the center of the circle?

Answer 1

If the circle has a radius of 15, then the diameter is 30.
No chord of the circle can be greater than the diameter.

Answer 2

Let $O$ be the center of the circle. Then $OA = OB = 15,$ so triangle $OAB$ is isosceles. It follows that segment $\overline{OP}$ is perpendicular to $\overline{AB},$ which means triangle $OPA$ is a right triangle. Taking $P$ to be the origin for convenience, we can then plot the points as shown below: [asy]
defaultpen(fontsize(10));

pair A_1, A_2, O, P;

O = (0,0);
A_1 = (15,0);
A_2 = (-15,0);
P = (0,-8);

draw(A_1--A_2);
draw(Circle(O,15));
draw(O--P);

label("$15$", (O + A_1)/2, S);
label("$15$", (O + A_2)/2, S);
label("$40$", (A_1 + A_2)/2, N);
label("$O$", O, NE);
label("$A$", A_1, dir(0));
label("$B$", A_2, dir(180));
label("$P$", P, SW);
[/asy] The length of $\overline{OP}$ is $15$ (the radius). We have that $POA$ and $POB$ are congruent right triangles, so $\overline{OP}$ bisects $\overline{AB}$. That means each half of $\overline{AB}$ is $20$, so $\overline{AB}$ would have to be $40 + 20 + 20 = \boxed{80}$ in order for $\overline{OP}$ to bisect it.

Answer 3

You are correct. My apologies for the error. No chord of the circle can be greater than the diameter, so a chord of length 40 is not possible. Therefore, there is no distance between the chord and the center of the circle.