Problem 63 - Intermediate Algebra

Question

By expanding \((x-h)^{2}+(y-k)^{2}=r^{2}\), we obtain \(x^{2}-\) \(2 h x+h^{2}+y^{2}-2 k y+k^{2}-r^{2}=0\). When we compare this result to the form \(x^{2}+y^{2}+D x+E y+F=0\), we see that \(D=-2 h, E=-2 k\), and \(F=h^{2}+k^{2}-r^{2}\). Therefore, the center and length of a radius of a circle can be found by using \(h=\frac{D}{-2}, k=\frac{E}{-2}\), and \(r=\sqrt{h^{2}+k^{2}-F}\). Use these relationships to find the center and the length of a radius of each of the following circles. (a) \(x^{2}+y^{2}-2 x-8 y+8=0\) (b) \(x^{2}+y^{2}+4 x-14 y+49=0\) (c) \(x^{2}+y^{2}+12 x+8 y-12=0\) (d) \(x^{2}+y^{2}-16 x+20 y+115=0\) (e) \(x^{2}+y^{2}-12 y-45=0\) (f) \(x^{2}+y^{2}+14 x=0\)

Step-by-Step Solution

Answer

(a) Center: (1, 4), Radius: 3; (b) Center: (-2, 7), Radius: 2; (c) Center: (-6, -4), Radius: 8; (d) Center: (8, -10), Radius: 7; (e) Center: (0, 6), Radius: 9; (f) Center: (-7, 0), Radius: 7.

Step 1: Identify coefficients D, E, and F for (a)

For equation (a): \(x^{2}+y^{2}-2x-8y+8=0\).- Identify - \(D = -2\) - \(E = -8\) - \(F = 8\)

Step 2: Calculate h, k, and r for (a)

For equation (a), use:\[h = \frac{D}{-2} = \frac{-2}{-2} = 1\]\[k = \frac{E}{-2} = \frac{-8}{-2} = 4\]\[r = \sqrt{h^{2} + k^{2} - F} = \sqrt{1^{2} + 4^{2} - 8} = \sqrt{9} = 3\]Center is (1, 4) and radius is 3.

Step 3: Identify coefficients D, E, and F for (b)

For equation (b): \(x^{2}+y^{2}+4x-14y+49=0\).- Identify - \(D = 4\) - \(E = -14\) - \(F = 49\)

Step 4: Calculate h, k, and r for (b)

For equation (b), use:\[h = \frac{D}{-2} = \frac{4}{-2} = -2\]\[k = \frac{E}{-2} = \frac{-14}{-2} = 7\]\[r = \sqrt{h^{2} + k^{2} - F} = \sqrt{(-2)^{2} + 7^{2} - 49} = \sqrt{4 + 49 - 49} = \sqrt{4} = 2\]Center is (-2, 7) and radius is 2.

Step 5: Identify coefficients D, E, and F for (c)

For equation (c): \(x^{2}+y^{2}+12x+8y-12=0\).- Identify - \(D = 12\) - \(E = 8\) - \(F = -12\)

Step 6: Calculate h, k, and r for (c)

For equation (c), use:\[h = \frac{D}{-2} = \frac{12}{-2} = -6\]\[k = \frac{E}{-2} = \frac{8}{-2} = -4\]\[r = \sqrt{h^{2} + k^{2} - F} = \sqrt{(-6)^{2} + (-4)^{2} + 12} = \sqrt{36 + 16 + 12} = \sqrt{64} = 8\]Center is (-6, -4) and radius is 8.

Step 7: Identify coefficients D, E, and F for (d)

For equation (d): \(x^{2}+y^{2}-16x+20y+115=0\).- Identify - \(D = -16\) - \(E = 20\) - \(F = 115\)

Step 8: Calculate h, k, and r for (d)

For equation (d), use:\[h = \frac{D}{-2} = \frac{-16}{-2} = 8\]\[k = \frac{E}{-2} = \frac{20}{-2} = -10\]\[r = \sqrt{h^{2} + k^{2} - F} = \sqrt{8^{2} + (-10)^{2} - 115} = \sqrt{64 + 100 - 115} = \sqrt{49} = 7\]Center is (8, -10) and radius is 7.

Step 9: Identify coefficients D, E, and F for (e)

For equation (e): \(x^{2}+y^{2}-12y-45=0\).- Identify - \(D = 0\) - \(E = -12\) - \(F = -45\)

Step 10: Calculate h, k, and r for (e)

For equation (e), use:\[h = \frac{D}{-2} = \frac{0}{-2} = 0\]\[k = \frac{E}{-2} = \frac{-12}{-2} = 6\]\[r = \sqrt{h^{2} + k^{2} - F} = \sqrt{0^{2} + 6^{2} + 45} = \sqrt{36 + 45} = \sqrt{81} = 9\]Center is (0, 6) and radius is 9.

Step 11: Identify coefficients D, E, and F for (f)

For equation (f): \(x^{2}+y^{2}+14x=0\).- Identify - \(D = 14\) - \(E = 0\) - \(F = 0\)

Step 12: Calculate h, k, and r for (f)

For equation (f), use:\[h = \frac{D}{-2} = \frac{14}{-2} = -7\]\[k = \frac{E}{-2} = \frac{0}{-2} = 0\]\[r = \sqrt{h^{2} + k^{2} - F} = \sqrt{(-7)^{2} + 0^{2} = 49} = 7\]Center is (-7, 0) and radius is 7.

Key Concepts

Completing the Square

Completing the square is a technique used to transform a quadratic equation into a perfect square trinomial. It is particularly helpful in converting the general form of a circle's equation into its standard form. This process involves rearranging and factoring terms in such a way that part of the equation becomes a squared binomial. Let's break it down into simple steps:

• Start with the quadratic term and the linear term. For example: If you have terms like \(x^2 + Bx\), your goal is to add and subtract the square of half of the coefficient of \(x\) to form a perfect square trinomial.
• For example, if you have \(x^2 + 4x\), half of 4 is 2, and 2 squared is 4. You would then rewrite \(x^2 + 4x\) as \((x + 2)^2 - 4\).
• This adjustment allows you to express the equation as a perfect square, facilitating the transition to a clearer form for further calculations.

Completing the square simplifies the process of identifying the circle's center and radius once equations are matched to their standard form counterpart.

Standard Form of a Circle

In the context of circle equations, the standard form is incredibly useful. It makes it easy to identify the circle’s center and radius directly by the equation format. The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\). Here's what each component represents:

• \(h\) and \(k\) are the x and y coordinates of the circle's center, respectively.
• \(r\) is the radius of the circle.

This form simplifies visualizing and graphing the circle and is reached by completing the square, if starting with the general form \(x^2 + y^2 + Dx + Ey + F = 0\). By comparing terms and rewriting the equation to match \((x-h)^2 + (y-k)^2 = r^2\), you can easily extract the circle's center and radius metrics from the equation. To convert a circle's equation in general form into this standard form, completing the square is a crucial intermediary step.

Radius and Center of a Circle

Finding the radius and center of a circle from its equation is straightforward once the equation is in standard form. Let's delve into how exactly you can derive these two important features: - **Center**: Given the standard form \((x-h)^2 + (y-k)^2 = r^2\), the center is easily identifiable as the point \((h, k)\). If given an equation in general form, like \(x^2 + y^2 + Dx + Ey + F = 0\), you can calculate \(h = \frac{D}{-2}\) and \(k = \frac{E}{-2}\). - **Radius**: The radius \(r\) can be found using the relationship \(r = \sqrt{h^2 + k^2 - F}\). It's crucial to understand how the radius originates from completing the square and converting the equation into its natural circular format. This ensures that all measures of the circle are captured accurately and graphically represented easily. Understanding how to extract these components based on transforming and comparing equations forms demonstrates the value of mathematical operations like completing the square and equation manipulation.