+195 The equation $x^2 + 6x + y^2 + 6y = 18$ represents a circle. What is its center? Write your answer as an ordered pair.
Find the $y$-intercept of the graph $x^2 - 10x + y^2 - 10y + 25 = 0$. Write your answer as an ordered pair.
Thanks in advance
x^2 + 6x + y^2 + 6y = 18 complete the square for x and y
(x+3)^2 + (y+3)^2 = 18 + 9 + 9
center h,k = -3,-3
x^2 -10x + y^2 -10y +25 = 0
(x-5)^2 + ( y-5)^2 = - 25 + 25 + 25 y intercept occurs when x = 0 put in x = 0
25 + y^2 -10y +25 = 25
y^2 -10y +25 = 0
(y-5)^2 = 0 y = 5 when x = 0 0,5
+79 You are right guest. Here is a more detailed ex
1. We complete the square twice. \begin{align*} x^2 + 6x + y^2 + 6y &= 18 \\ x^2 + 6x + 9+ y^2 + 6y +9 &= 18 +9+9\\ (x +3)^2 + (y + 3)^2 &= 36 \end{align*} This is in the form $(x - h)^2 + (y - k)^2 = r^2$, so the center is $\boxed{(-3, -3)}$.
2. We are given $x^2 - 10x + y^2 - 10y + 25 = 0$. Factoring, this is $x^2 - 10x + (y - 5)^2 = 0$. We use completing the square and add $25$ to both sides of the equation, giving us $x^2 - 10x +25 + (y - 5)^2 = 25$. Factoring again, this is $(x - 5)^2 + (y-5)^2 = 25$. This equation represents a circle of radius $5$, centered at $(5, 5)$. Thus, this circle will touch the $y$-axis at $\boxed{(0, 5)}$.
