Question

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 + square)^2 + (y + square)^2 = square ). the center of the circle is at the point ( ( square, square ) ).

Explanation:

Step1: Group x and y terms

$x^2 + 8x + y^2 + 22y = -37$

Step2: Complete the square for x

Take half of 8: $\frac{8}{2}=4$, square it: $4^2=16$. Add to both sides:
$x^2 + 8x + 16 + y^2 + 22y = -37 + 16$

Step3: Complete the square for y

Take half of 22: $\frac{22}{2}=11$, square it: $11^2=121$. Add to both sides:
$x^2 + 8x + 16 + y^2 + 22y + 121 = -37 + 16 + 121$

Step4: Rewrite as perfect squares

$(x+4)^2 + (y+11)^2 = 100$

Step5: Identify the center

For standard form $(x-h)^2+(y-k)^2=r^2$, center is $(h,k)$. Here $h=-4$, $k=-11$.

Answer:

Standard form: $(x + 4)^2 + (y + 11)^2 = 100$
Center: $(-4, -11)$