Question

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

Answer 1

Answer:

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

And solving we got:

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

If we subtract 100 from both sides we got:

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

$C= (-2,-10)$

$r = \sqrt{4}=2$

Step-by-step 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:

$r = \sqrt{4}=2$