Question

8 mark for review
$x^2 + y^2 + 6x + 8y - 11 = 0$
in the $xy$-plane, the graph of the given equation is a circle. what are the coordinates $(x, y)$ of the center of the circle?
a $(9, 16)$
b $(6, 8)$
c $(0, -11)$
d $(-3, -4)$

Explanation:

Step1: Recall the standard form of a circle's equation

The standard form of the equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius. To convert the given equation \(x^2 + y^2 + 6x + 8y - 11 = 0\) to this form, we complete the square for the \(x\)-terms and \(y\)-terms.

Step2: Group \(x\)-terms and \(y\)-terms

Group the \(x\)-terms and \(y\)-terms together: \((x^2 + 6x) + (y^2 + 8y) = 11\).

Step3: Complete the square for \(x\)-terms

For the \(x\)-terms \(x^2 + 6x\), take half of the coefficient of \(x\) (which is \(6\)), so \(\frac{6}{2}=3\), and square it: \(3^2 = 9\). Add this to both sides of the equation: \((x^2 + 6x + 9) + (y^2 + 8y) = 11 + 9\).

Step4: Complete the square for \(y\)-terms

For the \(y\)-terms \(y^2 + 8y\), take half of the coefficient of \(y\) (which is \(8\)), so \(\frac{8}{2}=4\), and square it: \(4^2 = 16\). Add this to both sides of the equation: \((x^2 + 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16\).

Step5: Rewrite as perfect squares

Now, rewrite the left side as perfect squares: \((x + 3)^2 + (y + 4)^2 = 36\). Comparing this with the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can see that \(h = -3\) and \(k = -4\) (since \((x + 3)^2=(x - (-3))^2\) and \((y + 4)^2=(y - (-4))^2\)).

Answer:

D. \((-3, -4)\)