Question

The equation of a circle in general form is

Answer 1

Answer:

• (x + 10)² + (y+6)²  = 121

Step-by-step explanation:

For this case we have the following expression:

$x^2 + y^2 +20 x + 12y +15=0$

And we want to write this on this general way:

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

So on this case we need to complete the squares like this:

$x^2 + 20 x + (\frac{20}{2})^2 + y^2 + 12y + (\frac{12}{2})^2 +15 =(\frac{20}{2})^2+(\frac{12}{2})^2$

Now we can subtract from both sides 15 and we got:

$(x^2 + 20 x +100) + (y^2 + 12y +36) = 100+36 - 15$

$(x+10)^2 +(y+6)^2 = 121$

And we can write the last expression like this:

$(x-(-10))^2 +(y-(-6))^2 = 121$

And if we compare to the general expression we see that:

$h = -10 , k = -6, r=\sqrt{121}=11$

So the correct option for this case would be:

• (x + 10)² + (y+6)²  = 121

Answer 2

Answer:

The answer to this would be your first option!!

• (x + 10)² + (y+6)2 = 121

Hope this helped, have a blessed day!!