Questions: Write the equation in standard form to find the center and radius of the circle. x^2 + y^2 + 6x + 14y + 37 = 0

Solution Steps

To find the center and radius of the circle given by the equation \(x^2 + y^2 + 6x + 14y + 37 = 0\), 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.

The given equation of the circle is: \[ x^2 + y^2 + 6x + 14y + 37 = 0 \]

First, we group the \(x\) terms together and the \(y\) terms together: \[ (x^2 + 6x) + (y^2 + 14y) + 37 = 0 \]

To complete the square for \(x\), we take half of the coefficient of \(x\), square it, and add and subtract it inside the equation: \[ x^2 + 6x = (x + 3)^2 - 9 \]

Similarly, to complete the square for \(y\), we take half of the coefficient of \(y\), square it, and add and subtract it inside the equation: \[ y^2 + 14y = (y + 7)^2 - 49 \]

Substitute the completed squares back into the equation: \[ (x + 3)^2 - 9 + (y + 7)^2 - 49 + 37 = 0 \]

Combine the constants: \[ (x + 3)^2 + (y + 7)^2 - 21 = 0 \]

Move the constant term to the other side of the equation: \[ (x + 3)^2 + (y + 7)^2 = 21 \]

The standard form of a circle's equation is: \[ (x - h)^2 + (y - k)^2 = r^2 \]

From the equation \((x + 3)^2 + (y + 7)^2 = 21\), we can identify:

• The center \((h, k)\) is \((-3, -7)\)
• The radius \(r\) is \(\sqrt{21}\)

Final Answer