Answer:
[tex] x^2 +4x +(4/2)^2 +y^2 -20 y +(20/2)^2 +100 = (4/2)^2 +(20/2)^2 [/tex]
And solving we got:
[tex] x^2 +4x+ 4 +y^2 +20y +100 +100 = 4 +100[/tex]
If we subtract 100 from both sides we got:
[tex] (x+2)^2 +(y+10)^2 = 4[/tex]
[tex] C= (-2,-10)[/tex]
[tex]r = \sqrt{4}=2[/tex]
Step-by-step explanation:
For this case we have the following equation given:
[tex] x^2 +4x +y^2 -20 y +100=0[/tex]
We want to find a general expression given by:
[tex] (x-h)^2 +(y-k)^2 = r^2[/tex]
Where the center is (h.k) and the radius is r.
We can begin completing the squares of the equation given and we got:
[tex] x^2 +4x +(4/2)^2 +y^2 -20 y +(20/2)^2 +100 = (4/2)^2 +(20/2)^2 [/tex]
And solving we got:
[tex] x^2 +4x+ 4 +y^2 +20y +100 +100 = 4 +100[/tex]
If we subtract 100 from both sides we got:
[tex] (x+2)^2 +(y+10)^2 = 4[/tex]
Then we can conclude that the center is:
[tex] C= (-2,-10)[/tex]
And the radius would be:
[tex]r = \sqrt{4}=2[/tex]
