Adiameter of a circle had 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

Center = -3,2

radius      = sqrt(65)

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

Step-by-step explanation:

We can find the center of the circle by finding the midpoint of the diameter

midpoint = (x1+x2)/2, (y1+y2)/2

              = (-10+4)/2, (-2+6)/2

               = (-6/2), (4/2)

                 = -3, 2

Center = -3,2  = (h,k)

The radius is the distance from the midpoint to one of the points on the diameter.  Using the points (-3,2) and (4,6)

distance = sqrt ( (x2-x1)^2 + (y1-y2)^2)

              = sqrt(( 4--3)^2 + (6 -2)^2)

             = sqrt(( 4+3)^2 + (6 -2)^2)

               = sqrt( 7^2 + 4^2)

               = sqrt(49+16)

radius      = sqrt(65)

The equation of a circle is given by

(x-h)^2 + (y-k) ^2 = r^2

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

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

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72

(-5,-28)

i graphed to on desmos.