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

We start with the general form of the circle's equation: \(x^{2}+y^{2}+8x + 22y+37 = 0\). Group the \(x\)-terms and \(y\)-terms together: \((x^{2}+8x)+(y^{2}+22y)= - 37\).

Step2: Complete the square for x

For the \(x\)-terms \(x^{2}+8x\), we take half of the coefficient of \(x\) (which is \(8\)), so \(\frac{8}{2}=4\), and square it: \(4^{2} = 16\). Add \(16\) to both sides of the equation.

Step3: Complete the square for y

For the \(y\)-terms \(y^{2}+22y\), we take half of the coefficient of \(y\) (which is \(22\)), so \(\frac{22}{2}=11\), and square it: \(11^{2}=121\). Add \(121\) to both sides of the equation.

Step4: Rewrite as perfect squares

After adding \(16\) and \(121\) to both sides, we have: \((x^{2}+8x + 16)+(y^{2}+22y+121)=- 37+16 + 121\). Now, rewrite the left - hand side as perfect squares: \((x + 4)^{2}+(y + 11)^{2}=100\).

The standard form of a circle's equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center of the circle. Comparing \((x + 4)^{2}+(y + 11)^{2}=100\) with the standard form, we can see that \(h=-4\) and \(k = - 11\), so the center of the circle is \((-4,-11)\).

Answer:

The equation of this circle in standard form is \((x + \boldsymbol{4})^{2}+(y + \boldsymbol{11})^{2}=\boldsymbol{100}\). The center of the circle is at the point \((\boldsymbol{-4},\boldsymbol{-11})\).