Question

Find the center \((h, k)\) and radius \(r\) of each circle; \((b)\) graph each circle; \((c)\) find the intercepts, if any. $$ x^{2}+y^{2}+4 x-4 y-1=0 $$

Short answer

Center: \(-2, 2\); Radius: \3\; x-intercepts: \(-2 + \sqrt{5}, 0\)\ and \(-2 - \sqrt{5}, 0\); y-intercepts: \(0, 2 + \sqrt{5}\)\ and \(0, 2 - \sqrt{5}\).

Step-by-step solution

- Group and Complete the Square

Group the x and y terms together, and move the constant term to the other side of the equation:\[ x^2 + 4x + y^2 - 4y = 1 \]To complete the square:1. For the x terms: Take half of 4, which is 2, then square it to get 4. Add and subtract 4 inside the equation.2. For the y terms: Take half of -4, which is -2, then square it to get 4. Add and subtract 4 inside the equation.This gives us:\[ (x^2 + 4x + 4) + (y^2 - 4y + 4) = 1 + 4 + 4 \]

- Simplify and Write in Standard Form

Now, simplify the left-hand side by writing it as squared terms:\[ (x+2)^2 + (y-2)^2 = 9 \]The circle equation is now in the standard form \((x-h)^2 + (y-k)^2 = r^2\), where \(h, k\) is the center and r is the radius.

- Identify the Center and Radius

By comparing \((x+2)^2 + (y-2)^2 = 9\) to the standard form:The center \(h, k\) is \(-2, 2\)\.The radius \(r\) is \(\sqrt{9} \) = \3\.

- Graph the Circle

Plot the center at (-2, 2), then draw a circle with a radius of 3 units around this center.

- Find the Intercepts

For the x-intercepts, set \(y = 0\) in \((x+2)^2 + (y-2)^2 = 9\):\[ (x+2)^2 + (0-2)^2 = 9 \]\[ (x+2)^2 + 4 = 9 \]\[ (x+2)^2 = 5 \]\[ x+2 = \pm\ \sqrt{5} \]\[ x = -2 \pm\ \sqrt{5} \]Thus, the x-intercepts are \(-2 + \sqrt{5}, 0\)\ and \(-2 - \sqrt{5}, 0\)\.For the y-intercepts, set \(x = 0\) in \((x+2)^2 + (y-2)^2 = 9\):\[ (0+2)^2 + (y-2)^2 = 9 \]\[ 4 + (y-2)^2 = 9 \]\[ (y-2)^2 = 5 \]\[ y-2 = .\pm\ \sqrt{5} \]\[ y = 2 \pm\ \sqrt{5} \]Thus, the y-intercepts are \(0, 2 + \sqrt{5}\) and \(0, 2 - \sqrt{5}\).

Key concepts

completing the square

Completing the square is a technique used to simplify quadratic expressions and is essential for rewriting the equation of a circle in its standard form.

To complete the square, we group the x-terms and y-terms separately and work on each group to form perfect squares. For example, let's start with the equation:
\[ x^2 + y^2 + 4x - 4y - 1 = 0 \]

• Group the x and y terms together: \[ x^2 + 4x + y^2 - 4y = 1 \]
• Take half the coefficient of x, which is 2, and square it to get 4. Add and subtract 4 within the equation.
• Do the same for y terms. Half of -4 is -2, and squaring it gives 4.

This results in:
\[ (x^2 + 4x + 4) + (y^2 - 4y + 4) = 1 + 4 + 4 \]

By rewriting the equation as a sum of squared terms, you get the equation in the standard form required to identify the center and the radius.

circle graphing

Graphing a circle involves plotting its center on the coordinate plane and drawing a circle with the radius around this point.

In our example, after completing the square, we simplified the equation to:
\[ (x+2)^2 + (y-2)^2 = 9 \]

Here, \((h, k) = (-2, 2)\) is the center, and the radius \(r = \sqrt{9} = 3\). To graph the circle:

• Plot the center at (-2, 2).
• With a radius of 3 units, measure 3 units in all directions (up, down, left, and right) from the center.
• Draw a smooth curve connecting these points to form the circle.

Thus, you get a perfect circle centered at (-2, 2) with a radius of 3 units.

finding intercepts

Finding the intercepts of a circle involves solving for the points where the circle intersects the x-axis and y-axis. These are called x-intercepts and y-intercepts.

To find the x-intercepts, set \(y = 0\) in the equation \((x+2)^2 + (y-2)^2 = 9\):
\[ (x+2)^2 + (0-2)^2 = 9 \]
Simplify to:
\[ (x+2)^2 + 4 = 9 \]
\[ (x+2)^2 = 5 \]
\[ x+2 = \pm\sqrt{5} \]
\[ x = -2 \pm\sqrt{5} \]

Thus, the x-intercepts are \((-2 + \sqrt{5}, 0)\) and \((-2 - \sqrt{5}, 0)\).

To find the y-intercepts, set \(x = 0\) in the equation:
\[ (0+2)^2 + (y-2)^2 = 9 \]
Simplify to:
\[ 4 + (y-2)^2 = 9 \]
\[ (y-2)^2 = 5 \]
\[ y-2 = \pm\root{5} \]
\[ y = 2 \pm\root{5} \]

Thus, the y-intercepts are \((0, 2 + \root{5})\) and \((0, 2 - \root{5})\).

Finding these intercepts helps understand where the circle crosses the axes, which is useful for graphing and analyzing the circle.