I have to find the center and the length of a radius of the circle:

Question

X^2+Y^2-2x-8y+8=0
Thanks,

Answer 1

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

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

c = 8

2g = -2

Divide each side by 2.

g  = -1

2f  = -8

Divide each side by 2.

f  = -4

Center (-g, -f ) = ( 1 , 4)

Radius(r ) = √(g ^2 + f ^2 - c )

= √(1 + 16 - 8)

= √ 9

= 3

Given circle center = ( 1 , 4)

r = 3.

Comments

Thank you :)

Answer 2

The standard form of the circle equation is ( x - h )² + ( y - k )² = r², where, (h, k) is the center of the circle, and r is the radius.

The equation is x² + y² - 2x - 8y + 8 = 0.

Write the equation in standard form of a circle.

To change the expression into a perfect square  add (half the x coefficient)² and add (half the y coefficient)²to each side of the expression.

Here, x coefficient = - 2, so, (half the x coefficient)² = (- 2/2)²= 1.

Here, y coefficient = - 8, so, (half the y coefficient)² = (- 8/2)²= 16.

Add 1 and 16 to each side.

x² - 2x + 1 + y² - 8y + 16 + 8 = 0 + 1 + 16

(x - 1)² + (y - 4)² = 17 - 8 = 9

(x - 1)² + (y - 4)² = 3² .

Compare the equation with standard form of a circle equation.

The center (h, k) is (1, 4), and

The radius (r) is 3 units.