The equation for the circle is:

x2+y2+14x+10y−7=0 .

What is the center of the circle?

Acceptfully
May 29, 2017

#1**+3 **

We want to get the equation of the circle into this form:

(x - h)^{2} + (y - k)^{2} = r^{2} , where (h, k) is the center and r is the radius.

x^{2} + y^{2} + 14x + 10y - 7 = 0

Subtract 7 from both sides of the equation and rearrange the left side.

x^{2} + 14x + y^{2} + 10y = 7

Add (14/2)^{2} , or 49, and (10/2)^{2} , or 25, to both sides of the equation.

x^{2} + 14x + 49 + y^{2} + 10y + 25 = 7 + 49 + 25

Now we can factor both parts like this

(x + 7)^{2} + (y + 5)^{2} = 81

Now that it is in that form, we can see that the center of the circle is (-7 , -5)

hectictar
May 29, 2017

#1**+3 **

Best Answer

We want to get the equation of the circle into this form:

(x - h)^{2} + (y - k)^{2} = r^{2} , where (h, k) is the center and r is the radius.

x^{2} + y^{2} + 14x + 10y - 7 = 0

Subtract 7 from both sides of the equation and rearrange the left side.

x^{2} + 14x + y^{2} + 10y = 7

Add (14/2)^{2} , or 49, and (10/2)^{2} , or 25, to both sides of the equation.

x^{2} + 14x + 49 + y^{2} + 10y + 25 = 7 + 49 + 25

Now we can factor both parts like this

(x + 7)^{2} + (y + 5)^{2} = 81

Now that it is in that form, we can see that the center of the circle is (-7 , -5)

hectictar
May 29, 2017

#2

#3**+2 **

In order to make x^{2} + 14x a perfect square trinomial, add (14/2)^{2}

x^{2} + 14x + (14/2)^{2}

x^{2} + 14x + 7^{2}

x^{2} + 14x + 49

Now it can be factored like this

(x + 7)(x + 7)

(x + 7)^{2}

And whatever you do to one side of the equation, do the same thing to the other side.

So we had to add 49 to both sides. This video might help explain it more:

hectictar
May 29, 2017