Question

3x² + 3y² - 24x - 12y + 33 = 0 is the equation of a circle with center (h, k) and radius r for: h = and k = and r = question help: video written example

Explanation:

Step1: Divide by 3 to simplify

Divide the entire equation \(3x^{2}+3y^{2}-24x - 12y + 33 = 0\) by 3. We get \(x^{2}+y^{2}-8x - 4y + 11 = 0\).

Step2: Group x and y terms

Group the \(x\)-terms and \(y\)-terms: \((x^{2}-8x)+(y^{2}-4y)= - 11\).

Step3: Complete the square for x

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

Step4: Complete the square for y

For the \(y\)-terms, take half of -4, which is -2, square it: \((-2)^{2}=4\). Add 4 to both sides.

Step5: Rewrite as perfect squares

The equation becomes \((x - 4)^{2}+(y - 2)^{2}=-11 + 16 + 4\).

Step6: Calculate the right side

Simplify the right side: \(-11 + 16 + 4 = 9\). So the equation is \((x - 4)^{2}+(y - 2)^{2}=3^{2}\).

Answer:

For the circle equation \((x - h)^{2}+(y - k)^{2}=r^{2}\), we have \(h = 4\), \(k = 2\), and \(r = 3\).

So, \(h=\boxed{4}\), \(k=\boxed{2}\), \(r=\boxed{3}\)