Questions: Question 14, 2.2 .98 a. Find the center and radius of the circle x^2 + y^2 + x = 0. b. Find the x-and y-intercepts of the graph of the circle x^2 + y^2 + x = 0.

Solution Steps

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

b. To find the \(x\)- and \(y\)-intercepts of the circle, we set \(y = 0\) to find the \(x\)-intercepts and set \(x = 0\) to find the \(y\)-intercepts.

Given the equation of the circle: \[ x^2 + y^2 + x = 0 \]

We need to rewrite it in the standard form \((x-h)^2 + (y-k)^2 = r^2\). To do this, we complete the square for the \(x\) terms.

Rewrite the equation: \[ x^2 + x + y^2 = 0 \]

Complete the square for the \(x\) terms: \[ x^2 + x = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} \]

So the equation becomes: \[ \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + y^2 = 0 \]

Rearrange to standard form: \[ \left(x + \frac{1}{2}\right)^2 + y^2 = \frac{1}{4} \]

From the standard form \((x-h)^2 + (y-k)^2 = r^2\), we can identify:

• Center: \((-h, -k) = \left(-\frac{1}{2}, 0\right)\)
• Radius: \(r = \sqrt{\frac{1}{4}} = \frac{1}{2}\)

To find the \(x\)-intercepts, set \(y = 0\) in the original equation: \[ x^2 + x = 0 \] \[ x(x + 1) = 0 \] So, the \(x\)-intercepts are: \[ x = 0 \quad \text{and} \quad x = -1 \]

To find the \(y\)-intercepts, set \(x = 0\) in the original equation: \[ y^2 = 0 \] So, the \(y\)-intercept is: \[ y = 0 \]

Final Answer