Question

a circle has the equation 3x² + 24x + 3y² = 0. (a) find the center (h,k) and radius r of the circle. (b) graph the circle. (c) find the intercepts, if any, of the graph.

Explanation:

Step1: Rewrite the given equation in standard form

First, divide the entire equation $3x^{2}+24x + 3y^{2}=0$ by 3 to get $x^{2}+8x + y^{2}=0$. Then complete the square for the $x$ - terms. The formula for completing the square for $x^{2}+bx$ is $(x+\frac{b}{2})^{2}-\frac{b^{2}}{4}$. For $x^{2}+8x$, we have $(x + 4)^{2}-16$. So the equation becomes $(x + 4)^{2}-16+y^{2}=0$, which can be rewritten as $(x + 4)^{2}+y^{2}=16$.

Step2: Identify the center and radius

The standard - form of a circle equation is $(x - h)^{2}+(y - k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle and $r$ is the radius. Comparing $(x + 4)^{2}+y^{2}=16$ with the standard form, we have $h=-4$, $k = 0$, and $r = 4$.

Step3: Find the x - intercepts

To find the x - intercepts, set $y = 0$ in the equation $(x + 4)^{2}+y^{2}=16$. Then $(x + 4)^{2}=16$. Taking the square root of both sides, we get $x+4=\pm4$. When $x + 4=4$, $x = 0$; when $x + 4=-4$, $x=-8$.

Step4: Find the y - intercepts

To find the y - intercepts, set $x = 0$ in the equation $(x + 4)^{2}+y^{2}=16$. Then $(0 + 4)^{2}+y^{2}=16$, which simplifies to $16+y^{2}=16$, so $y = 0$.

Answer:

(a) Center: $(-4,0)$, Radius: $r = 4$
(b) To graph the circle, plot the center at the point $(-4,0)$ and then draw a circle with a radius of 4 units around it.
(c) x - intercepts: $x = 0$ and $x=-8$; y - intercept: $y = 0$