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Chapter 13: Problem 14

Find the center and radius of each circle. (Hint: In Exercises 15 and \(16,\) divide each side by a common factor) $$ x^{2}+y^{2}-2 x+4 y-4=0 $$

Short Answer

Expert verified
Center: (1, -2), Radius: 3

Step by step solution

01

Rewrite the equation in standard form

Rewrite the given equation by moving the constant term to the right side of the equation: \[ x^{2} + y^{2} - 2x + 4y = 4 \]
02

Group the x and y terms together

Group the x terms together and the y terms together: \[ (x^{2} - 2x) + (y^{2} + 4y) = 4 \]
03

Complete the square for the x terms

To complete the square for the x terms, add and subtract \( \frac{(-2/2)^{2}}{1} \to (1) \) inside the parentheses: \[ (x^{2} - 2x + 1 - 1) + (y^{2} + 4y) = 4 \] This simplifies to: \[ (x - 1)^{2} - 1 + (y^{2} + 4y) = 4 \]
04

Complete the square for the y terms

To complete the square for the y terms, add and subtract \( \frac{(4/2)^{2}}{1} \to (4) \) inside the parentheses: \[ (x - 1)^{2} - 1 + (y^{2} + 4y + 4 - 4) = 4 \] This simplifies to: \[ (x - 1)^{2} - 1 + (y + 2)^{2} - 4 = 4 \]
05

Combine and simplify

Combine and simplify the equation by isolating the completed squares on one side and the constants on the other: \[ (x - 1)^{2} + (y + 2)^{2} - 1 - 4 = 4 \] This further simplifies to: \[ (x - 1)^{2} + (y + 2)^{2} = 9 \]
06

Identify the center and radius

Compare the simplified equation to the standard form of a circle \((x - h)^{2} + (y - k)^{2} = r^{2}\). Here, the center (h,k) is (1, -2) and the radius r is the square root of 9, which is 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

completing the square
Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This is particularly useful for rewriting the equation of a circle.
To complete the square for the expression involving x in the equation \(x^2 - 2x\):
  • Take the coefficient of x, which is -2.
  • Halve it to get -1.
  • Square -1 to get 1.
  • Add and subtract this square inside the equation: \((x^2 - 2x + 1 - 1)\).
This converts \(x^2 - 2x + 1\) into \((x - 1)^2\). Repeat the same process for the y terms in the equation. This allows us to rewrite the equation in a form that makes it easier to identify the circle's geometric properties.
standard form of a circle
The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \(h, k\) is the center of the circle and \(r\) is its radius.
In practice:
  • First, ensure the circle's equation is structured so that the constant term is on the right side, with x and y grouped together on the left.
  • Then, complete the square for both x and y terms to bring the equation into the form of \((x - h)^2 + (y - k)^2\).
  • The simplification of the left side should yield a perfect match to this standard form, resulting in a clear identification of the circle's center and radius.
Applying these steps helps transition from a general quadratic equation to the recognizable standard form, simplifying the extraction of key circle characteristics.
center and radius of a circle
After rewriting the circle's equation in standard form, identifying the center and radius becomes straightforward.
For example, with the equation \((x - 1)^2 + (y + 2)^2 = 9\):
  • Compare it with the standard form to see that \(h = 1\) and \(k = -2\).
  • Hence, the center of the circle is at \((1, -2)\).

  • The radius is derived from the value on the right side of the equation. Since \(9\) equals \(r^2\), taking the square root of 9 gives us \(r = 3\).
Understanding this connection allows you to quickly determine any circle's center and radius once its equation is in standard form.

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Most popular questions from this chapter

Solve each system by the substitution method. \(y=4 x^{2}-x\) \(y=x\)

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In rugby, after a try (similar to a touchdown in American football) the scoring team attempts a kick for extra points. The ball must be kicked from directly behind the point where the try was scored. The kicker can choose the distance but cannot move the ball sideways. It can be shown that the kicker's best choice is on the hyperbola with equation $$\frac{x^{2}}{g^{2}}-\frac{y^{2}}{g^{2}}=1$$ where \(2 g\) is the distance between the goal posts. since the hyperbola approaches its asymptotes, it is easier for the kicker to estimate points on the asymptotes instead of on the hyperbola. What are the asymptotes of this hyperbola? Why is it relatively easy to estimate them? (Source: Isaksen, Daniel C., "How to Kick a Field Goal," The College Mathematics Journal.)
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