center C has a radius of 13.Chord AB=10.find the distance from the center of the circle to chord AB. CD=?

make a sketch, join the centre to A, and complete the right-angled triangle.

let the distance from the centre to the chord be x
x^2 + 5^2 = 13^2
x^2 = 169-25 = 144
x = √144 = 12

Where does CD enter the picture?
Where is D ?
Your opening sentence of "center C has a radius of 13" makes no sense.

To find the distance from the center of the circle to chord AB, we need to draw perpendiculars from the center to chord AB. Let's call the point where the perpendicular intersects chord AB as point D.

Now, we can form a right-angled triangle with CD as the hypotenuse, and AD and BD as the other two sides. Since the perpendicular drawn from the center to a chord bisects the chord, we know that AD = BD = 5.

To find CD, we can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In this case, CD is the hypotenuse, so we can write:

CD^2 = AD^2 + BD^2

Substituting the values we know, we get:

CD^2 = 5^2 + 5^2
CD^2 = 25 + 25
CD^2 = 50

To find CD, we take the square root of both sides:

CD = sqrt(50)

So, the distance from the center of the circle to chord AB, CD, is equal to sqrt(50).