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