Question

Find the center and the radius of this equation by completing the

square.

For the values of x and y, be sure to include the "+" or "-" sign before

the number.

x2 + 4x + y2 – 20y + 100 = 0

C=( -2

10

Dr=

Answer 1

`x^2 +4x +(4/2)^2 +y^2 -20 y +(20/2)^2 +100 = (4/2)^2 +(20/2)^2`

And solving we got:

`x^2 +4x+ 4 +y^2 +20y +100 +100 = 4 +100`

If we subtract 100 from both sides we got:

`(x+2)^2 +(y+10)^2 = 4`

`C= (-2,-10)`

`r = √(4)=2`

Explanation:

For this case we have the following equation given:

`x^2 +4x +y^2 -20 y +100=0`

We want to find a general expression given by:

`(x-h)^2 +(y-k)^2 = r^2`

Where the center is (h.k) and the radius is r.

We can begin completing the squares of the equation given and we got:

`x^2 +4x +(4/2)^2 +y^2 -20 y +(20/2)^2 +100 = (4/2)^2 +(20/2)^2`

And solving we got:

`x^2 +4x+ 4 +y^2 +20y +100 +100 = 4 +100`

If we subtract 100 from both sides we got:

`(x+2)^2 +(y+10)^2 = 4`

Then we can conclude that the center is:

`C= (-2,-10)`

And the radius would be:

`r = √(4)=2`