algebra equation?

Question

express the equation of the circle in standard form. State center and radius

a.) x^2 = y^2 - 6x +10y +9 = 0

need help with this one broken down please =)

Answer 1

(FIrst off, you have too many equal signs. I'm going to assume the 1st one is supposed to by a plus.)

x² + y² - 6x +10y +9 = 0
1st, move all the variable terms to one side and the constants to the other:
x² + y² - 6x +10y = -9
Group the x terms and y terms together:
(x² -6x )+ (y² +10y ) = -9
Complete the square for the x terms and the y terms,
making sure to add the constants to both sides of the
equation:
(x² -6x +9)+ (y² +10y + 25 ) = -9 + 9 + 25
Rewrite the two trinomials as perfect square binomials:
(x-3)² + (y+5)² = 25
This is now in standard form: (x-h)² + (y-k)² = r²
Your h=3, k= - 5, and your r = 5
So this is a circle with its center at (3,-5) and a radius of 5.

Answer 2

x^2 = y^2 - 6x +10y +9 = 0

x² = 0

x = 0

y² +10y +9 = 0

(y + 1)(y + 9) = 0

y = -1, y = -9, x = 0
.