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