Type the correct answer in each box The general form of the equation of a circle is x^2+y^2+8x+22y+37=0

The equation of this circle in standard form is (x+__)^2+(y+__)^2=__. The center of the circle is at the point (__,__).

We know that the equation of the circle in standard form is equal to (x-h)² + (y-k)² = r² where (h,k) is the center of the circle and r is the radius of the circle.

We have x² + y² + 8x + 22y + 37 = 0, let's get to the standard form :

1 - We first group terms with the same variable :

(x²+8x) + (y²+22y) + 37 = 0

2 - We then move the constant to the opposite side of the equation (don't forget to change the sign !)

(x²+8x) + (y²+22y) = - 37

3 - Do you recall the quadratic identities ? (a+b)² = a² + 2ab + b². Now that's what we are trying to find. We call this process "completing the square".

x²+8x = (x²+8x + 4²) - 4² = (x+4)² - 4²

y²+22y = (y²+22y+11²)-11² = (y+11)²-11²

4 - We plug the new values inside our equation :

(x+4)² - 4² + (y+22)² - 11² = -37

(x+4)² + (y+22)² = -37+4²+11²

(x+4)²+(y+22)² = 100

5 - We re-write in standard form :

(x-(-4)²)² + (y - (-22))² = 10²

And now it is easy to identify h and k, h = -4 and k = - 22 and the radius r equal 10. You can now complete the sentence :)

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