If the diameter of a circle has endpoints A(7, 2) and B(-1, 8), where is the center of the circle?

Solution:

The diameter passes through the center of the circle and its endpoints are on the circumference of the circle.

Given the endpoints(x\(_1\), y\(_1\)) and (x\(_2\), y\(_2\)).

Using the midpoint formula

M (x\(_3\), y\(_3\)) = [(x\(_1\) + x\(_2\))/2, (y\(_1\) + y\(_2\))/2]

The two points given in the question are A(7, 2) and B(-1, 8)

Substituting it in the formula

M (x\(_3\), y\(_3\)) = [(7 - 1)/2, (2 + 8)/2]

By further calculation

M (x3, y3) = [6/2, 10/2]

So we get

M (x3, y3) = [3, 5]

Therefore, the center of the circle is (3, 5).

If the diameter of a circle has endpoints A(7, 2) and B(-1, 8), where is the center of the circle?

Summary:

If the diameter of a circle has endpoints A(7, 2) and B(-1, 8), the center of the circle is (3, 5).