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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

 Feb 10, 2021
 #1
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+1

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

 Feb 10, 2021
 #2
avatar+79 
+2

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 $\boxed{(0, 5)}$.

AnswerMachine  Feb 10, 2021

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