Question

The graph of the equation below is a circle. Identify the center and radius of the circle. x^2 + 10x + y^2 − 8y − 8 = 0

Answer 1

Answer:

$(x+5)^2 + (y-4)^2 = 49$

If we compare this equation with the general formula for a circle given by:

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

We see that h = -5 and k = 4 so then the center is V = (-5,4). And the radius would be $r = \sqrt{49}= 7$

Step-by-step explanation:

For this case we have the following expression:

$x^2 + 10 x + y^2 -8y -8 =0$

We can complete the squares for this case like this"

$[x^2 +10 x +(\frac{10}{2})^2] + [y^2 -8y +(\frac{8}{2})^2] = 8 +(\frac{10}{2})^2 + (\frac{8}{2})^2$

And we can express this like that:

$(x^2 +10x + 25) +(y^2 -8y + 16)= 8+ 25+ 16=49$

And we can simplify like this:

$(x+5)^2 + (y-4)^2 = 49$

If we compare this equation with the general formula for a circle given by:

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

We see that h = -5 and k = 4 so then the center is V = (-5,4). And the radius would be $r = \sqrt{49}= 7$