Adiameter of a circle has endpoints p (-10,-2) and q (4,6) a. find the center of the circle. b. find the radius. if your answer is not an integer, express it in radical form. c. write an equation for the circle.

For the answer to the question above, in this case, the center is 

((-10 + 4)/2, (-2 + 6)/2) = (-6/2, 4/2) = (-3,2). (To determine a midpoint you just 

take the average of the x-values for the x-coordinate and the average of the 

y values for the y-coordinate of the midpoint.) 

b. For the radius so 

r = (1/2) * sqrt((4 - (-10))^2 + (6 - (-2))^2) = (1/2) * sqrt(14^2 + 8^2) = (1/2)*sqrt(260) 

which can be simplified to (1/2)*2sqrt(65) = sqrt(65). 

(That is, r = (1/2) * diameter and the diameter is then calculated using the 

formula for the distance between the two endpoints given.) 

c. The general equation for a circle is (x-h)^2 + (y-k)^2 = r^2, center (h,k) radius r 

so in this case we have (x + 3)^2 + (y - 2)^2 = 65, as r^2 = (sqrt(65))^2 = 65.
I hope my answer helped you.

Letter a i would say