Question

3x² + 3y² - 24x - 30y - 69 = 0 is the equation of a circle with center (h,k) and radius r for: h = and k = and r = question help: video

Explanation:

Step1: Divide the equation by 3

Divide each term in \(3x^{2}+3y^{2}-24x - 30y-69 = 0\) by 3.
We get \(x^{2}+y^{2}-8x - 10y-23 = 0\).

Step2: Complete the square for \(x\) and \(y\) terms

For the \(x\) - terms: \(x^{2}-8x=(x - 4)^{2}-16\).
For the \(y\) - terms: \(y^{2}-10y=(y - 5)^{2}-25\).
Substitute these into the equation: \((x - 4)^{2}-16+(y - 5)^{2}-25-23 = 0\).

Step3: Simplify the equation

Simplify \((x - 4)^{2}+(y - 5)^{2}-16 - 25-23 = 0\) to \((x - 4)^{2}+(y - 5)^{2}=64\).

Answer:

\(h = 4\)
\(k = 5\)
\(r = 8\)