Question

Find the center \((h, k)\) and radius \(r\) of each circle; \((b)\) graph each circle; \((c)\) find the intercepts, if any. $$ 2 x^{2}+8 x+2 y^{2}=0 $$

Short answer

(h, k) = (-2, 0), r = 2. Intercepts: (0, 0), (-4, 0).

Step-by-step solution

- Rewrite the equation in a standard form

Divide every term by 2 to simplify: x^2 + 4x + y^2 = 0.

- Complete the square for the x-terms

Group the x terms and complete the square: x^2 + 4x = (x+2)^2 - 4. So, the equation becomes (x + 2)^2 - 4 + y^2 = 0.

- Isolate terms to form a circle equation

Move the constants to one side to form the standard circle equation: (x + 2)^2 + y^2 = 4.

- Identify the center and radius

By comparing to the standard form (x - h)^2 + (y - k)^2 = r^2, it is clear that h = -2, k = 0, and r^2 = 4. So, r = 2.

- Graph the circle

Plot the center (-2, 0) and use the radius 2 to draw the circle. The circle will have points 2 units away from the center in all directions.

- Find the intercepts

To find the x-intercepts set y = 0: (x + 2)^2 + 0^2 = 4, so, (x + 2)^2 = 4 x + 2 = ±2 x = 0 or x = -4. Intercepts are (0, 0) and (-4, 0). To find y-intercepts set x = 0: (0 + 2)^2 + y^2 = 4 4 + y^2 = 4 y^2 = 0 => y = 0. The intercept is (0, 0).

Key concepts

center of a circle

The center of a circle is a fixed point from which all points on the circle are equidistant. In the standard equation of a circle, (x - h)^2 + (y - k)^2 = r^2, the coordinates (h, k) represent the center.When we look at the given equation 2x^2 + 8x + 2y^2 = 0, it is initially not in standard form.
To convert it, we first simplify by dividing all terms by 2: x^2 + 4x + y^2 = 0.
Next, we complete the square for the x-terms. Group the x-terms and reformat: x^2 + 4x = (x + 2)^2 - 4.
Rewriting the equation incorporating this, we get: (x + 2)^2 - 4 + y^2 = 0.
Moving constants to the other side provides: (x + 2)^2 + y^2 = 4.
Now, we can see that the equation matches our standard form with (h, k) = (-2, 0). So, the center of the circle is (-2, 0).

radius of a circle

The radius of a circle is the distance from its center to any point on the circle. In the standard circle equation (x - h)^2 + (y - k)^2 = r^2, the radius is given by r. In our simplified equation: (x + 2)^2 + y^2 = 4, we can see that this matches the standard circle form with 4 as the value of r^2.
Taking the square root of both sides, r^2 = 4 implies r = 2.
Therefore, the radius of the circle is 2 units.

completing the square

Completing the square is a method used to convert a quadratic equation into a standard form. This makes it easier to identify key attributes like the center and radius of a circle.To complete the square for the given equation, we start with x^2 + 4x + y^2 = 0.First, we focus on the x-terms: x^2 + 4x.
To create a perfect square trinomial, we take half the coefficient of x (which is 4), square it to get 4, and add & subtract this value within the equation. So, we reformat x^2 + 4x as (x + 2)^2 - 4.The equation now becomes: (x + 2)^2 - 4 + y^2 = 0.
Next, we isolate the terms to form the standard circle equation: (x + 2)^2 + y^2 = 4.Now, this equation matches the standard form of a circle equation, making it easier to read the center and radius.

graphing circles

Graphing circles involves plotting the circle's center and using its radius to draw the circle. Using our equation, (x + 2)^2 + y^2 = 4, we identified the center (-2, 0) and radius 2 units.To graph it:

• First, plot the center of the circle at (-2, 0).
• From the center, measure 2 units outwards in all directions.

These points mark the boundary of the circle.Connect these points smoothly in a circular shape.
The result is a circle centered at (-2, 0) with all points on the circle 2 units away from the center.

intercepts of a circle

Intercepts are points where the circle crosses the x-axis or y-axis.For x-intercepts, set y = 0 in the simplified equation (x + 2)^2 + y^2 = 4.Replacing y, we get: (x + 2)^2 + 0 = 4,which leads to (x + 2)^2 = 4.Solving this: x + 2 = ±2 gives two values: x = 0 or x = -4.
The x-intercepts are thus (0, 0) and (-4, 0).For y-intercepts, set x = 0 in the equation: (0 + 2)^2 + y^2 = 4.This simplifies to 4 + y^2 = 4.Subtracting 4 from both sides: y^2 = 0,leads to y = 0.
Therefore, the y-intercept is (0, 0).
Hence, our circle's intercepts are (0, 0) and (-4, 0).