Questions: 2x^2 + 2y^2 + 20x + 12y - 60 = 0 is the equation of a circle with center (h, k) and radius r for: h= and k= and r=

Solution Steps

To find the center \((h, k)\) and radius \(r\) of the circle given by the equation \(2x^2 + 2y^2 + 20x + 12y - 60 = 0\), we need to rewrite it in the standard form \((x-h)^2 + (y-k)^2 = r^2\). This involves completing the square for both \(x\) and \(y\) terms.

Start with the given equation of the circle: \[ 2x^2 + 2y^2 + 20x + 12y - 60 = 0 \]

Divide the entire equation by 2 to simplify: \[ x^2 + y^2 + 10x + 6y - 30 = 0 \]

For the \(x\) terms, complete the square: \[ x^2 + 10x \] \[ = (x + 5)^2 - 25 \]

For the \(y\) terms, complete the square: \[ y^2 + 6y \] \[ = (y + 3)^2 - 9 \]

Substitute the completed squares back into the equation: \[ (x + 5)^2 - 25 + (y + 3)^2 - 9 - 30 = 0 \]

Simplify: \[ (x + 5)^2 + (y + 3)^2 = 64 \]

The equation \((x + 5)^2 + (y + 3)^2 = 64\) is in the standard form \((x-h)^2 + (y-k)^2 = r^2\).

• Center \((h, k)\) is \((-5, -3)\).
• Radius \(r\) is \(\sqrt{64} = 8\).

Final Answer