Question

A diameter of a circle has endpoints P(-10,-2) and Q(4,6)

Answer 1

A. The centre must be the midpoint of PQ 
which would be C( (-10+4)/2 , (-2+6)/2 ) 
= C(-3,2) 

B. radius is √( (4+3)^2 + (2-6)^2) = √65 

C. equation: 

(x+3)^2 + (y-2)^2 = 65

Answer 2

A. Find the center of the circle.
 Using the middle point formula we have:
 C = ((x1 + x2) / 2, (y1 + y2) / 2)
 Substituting values:
 C = ((- 10 + 4) / 2, (-2 + 6) / 2)
 C = ((- 6) / 2, (4) / 2)
 C = (- 3, 2)

 B. FInd the radius. If your answer is not an integer, express it in radical form

 Using the formula of distance between points we have:
 d = root ((x2-x1) ^ 2 + (y2-y1) ^ 2)
 Substituting values:
 d = root ((4 - (- 10)) ^ 2 + (6 - (- 2)) ^ 2)
 d = root ((4 + 10) ^ 2 + (6 + 2) ^ 2)
 d = root ((14) ^ 2 + (8) ^ 2)
 d = root (260)
 d = 2 * root (65)
 Then, the radius of the circle is:
 r = d / 2
 r = (2 * root (65)) / 2
 r = root (65)

 C write an equation for the circle

 The equation of the circle is:
 (x + 3) ^ 2 + (y-2) ^ 2 = 65