Write the standard form of the equation of the circle with the given characteristics

Question

Write the standard form of the equation of the circle with the given characteristics....Endpoints of a diameter: (2, 3), (−8, −9)

Answer 1

the center is at (-3, -3) which is the midpoint

the radius, per the distance formula is sqrt ( ( -3 - -8)^2 + (-3 - -9)^2 )

= sqrt( 5^2 + 6^2)

= sqrt( 25 + 36)

= sqrt(61)

(x+3)^2 + (y+3)^2 = 61

Answer 2

First, find the center point of the diameter of the circle: (average each coordinate direction)

the x axis centerpoint = [2 + (-8)] / 2 = -6/2 = -3

the y axis centerpoint = [3 + (-9)] / 2 = -6/2 = -3

therefore the center of the circle is (-3,-3) or (h,k)

therefore, writing the standard circle equation:

(x – h)^2 + (y – k)^2 = r^2

substitute:

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

(x + 3)^2 + (y + 3)^2 = r^2

now to find the radius, need to plug one of the diameter endpoints and solve for r:

@ x=2, y=3

(2 + 3)^2 + (3 + 3)^2 = r^2

5^2 + 6^2 = r^2

25 + 36 = r^2

61 = r^2

r = squareroot (61) = 7.81

complete equation: (x + 3)^2 + (y + 3)^2 = (7.81)^2