Question

what is the center for the circle with this equation $x^2 + y^2 + 4x + 6y - 36 = 0$? (use desmos)
(-2,5)
(-2,-3)
(2,7)

Explanation:

Step1: Rewrite the circle equation

We start with the general form of the circle equation \(x^{2}+y^{2}+4x + 6y-36 = 0\). We complete the square for the \(x\) and \(y\) terms separately.

For the \(x\) - terms: \(x^{2}+4x=(x + 2)^{2}-4\) (using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\) and \(2ab = 4x\), so \(b = 2\))

For the \(y\) - terms: \(y^{2}+6y=(y + 3)^{2}-9\) (using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a = y\) and \(2ab=6y\), so \(b = 3\))

Substitute these back into the original equation:

\((x + 2)^{2}-4+(y + 3)^{2}-9-36=0\)

Step2: Simplify the equation

Simplify the left - hand side of the equation:

\((x + 2)^{2}+(y + 3)^{2}-4-9 - 36=0\)

\((x + 2)^{2}+(y + 3)^{2}-49 = 0\)

Then, \((x + 2)^{2}+(y + 3)^{2}=49\)

The standard form of a circle equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center of the circle and \(r\) is the radius.

Comparing \((x + 2)^{2}+(y + 3)^{2}=49\) with \((x - h)^{2}+(y - k)^{2}=r^{2}\), we have \(h=-2\) and \(k=-3\)

Answer:

B. \((-2,-3)\)