please help!

Question

(1) find the center of the circle whose equation is

x^2+ y^2-3x+8y-8=0

(2) write the equation of the circle with center (-2,1) and which passes through (3,4).

Answer 1

1) Given circle equation x ^2 + y ^2 -3x + 8y - 8 = 0

x ^2 + y ^2 - 3x + 8y + 8  = 0

Compare it to circle equation x ^2 + y ^2 + 2gx + 2fy + c  = 0.

c = -8

2g = -3

Divide each side by 2.

g  = -3/2

2f  = 8

Divide each side by 2.

f  = 4

Center (-g, -f ) = ( 3/2 , -4)

Radius(r ) = √(g ^2 + f ^2 - c )

= √((9/4) + 16 - (-8))

= √ ((9/4) + 16 + 8)

= √ ((9/4) + 24)

= √((9+96)/4)

= √(105/4)

= √26.25

= 5.12

Given circle center = ( 3/2 , -4)

r = 5.12.

Answer 2

2) Circle center (h, k ) = (- 2, 1) and the point is (3, 4).

The distance between center and the point is would be the radius of circle.

Radius (r ) = sqrt [(x -h )² + (y -k )²].

Substitute the values of  (h, k ) = (- 2, 1) and (x , y ) = (3, 4) in (r ) = sqrt [(x -h )² + (y -k )²].

r = sqrt [(3 + 2)² + (4 - 1)²]

= sqrt [(5)² + (3)²]

= sqrt [25 + 9 ]

= sqrt [34].

= 5.8

Circle equation is (x - h )² + (y - k )² = r ².

(x + 2 )² + (y - 1 )² = (5.8) ²

(x + 2)² + (y - 1 )² =  34

x ² + 4x + 4 + y ² - 2y + 1 = 34

x ² + 4x + y ² - 2y + 5 = 34

Subtract 34 from each side.

x ² + y ² + 4x - 2y + 5 - 34 = 0.

Circle equation is x ² + y ² + 4x - 2y - 29 = 0.