By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
The center-radius form and general form of a circle are two ways to express the equation of a circle in mathematics. The center-radius form, also known as the standard form, is a way to describe a circle by its center point and radius, while the general form is a more compact way to represent the equation of a circle.
Circles are essential in mathematics and have numerous real-world applications, such as in geometry, trigonometry, and engineering. In data analysis, circles are used to represent data points, trends, and patterns. In science, circles are used to model the orbits of celestial bodies, the shape of molecules, and the curvature of space-time. In engineering, circles are used to design curves, shapes, and structures.
The center-radius form of a circle is given by the equation:
$$(x - h)^2 + (y - k)^2 = r^2$$
where $(h, k)$ is the center of the circle and $r$ is the radius.
The general form of a circle is given by the equation:
$$Ax^2 + Ay^2 + Bx + Cy + D = 0$$
where $A$, $B$, $C$, and $D$ are constants.
To convert the general form to the center-radius form, we need to complete the square. This involves rewriting the equation in a way that allows us to easily identify the center and radius of the circle.
To solve a problem involving circles, we need to identify the center and radius of the circle. This can be done by looking at the center-radius form of the equation and identifying the values of $h$, $k$, and $r$.
If we are given the center-radius form of the equation, we may need to convert it to the general form to solve the problem. This can be done by expanding the squared terms and rearranging the equation.
If we are given the general form of the equation, we need to complete the square to convert it to the center-radius form. This involves rewriting the equation in a way that allows us to easily identify the center and radius of the circle.
Once we have the equation in the center-radius form, we can easily identify the center and radius of the circle. We can then use this information to solve the problem.
Problem Statement:
$$x^2 + y^2 - 6x + 4y - 12 = 0$$
Solution:
$$x^2 - 6x + 9 + y^2 + 4y + 4 = 12 + 9 + 4$$
$$(x - 3)^2 + (y + 2)^2 = 25$$
Answer:
The center of the circle is $(3, -2)$ and the radius is $5$.
$$x^2 + 4x + y^2 - 6y = 0$$
$$x^2 + 4x + 4 + y^2 - 6y + 9 = 4 + 9$$
$$(x + 2)^2 + (y - 3)^2 = 13$$
The center of the circle is $(-2, 3)$ and the radius is $\sqrt{13}$.
$$(x - 2)^2 + (y + 1)^2 = 16$$
The center of the circle is $(2, -1)$ and the radius is $4$.
What is the center-radius form of a circle?
A) $(x - h)^2 + (y - k)^2 = r^2$ B) $Ax^2 + Ay^2 + Bx + Cy + D = 0$ C) $(x + h)^2 + (y + k)^2 = r^2$ D) $(x - h)^2 - (y - k)^2 = r^2$
Correct Answer: A Explanation: The center-radius form of a circle is given by the equation $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius.
What is the general form of a circle?
Correct Answer: B Explanation: The general form of a circle is given by the equation $Ax^2 + Ay^2 + Bx + Cy + D = 0$, where $A$, $B$, $C$, and $D$ are constants.
What is the center of a circle with the equation $(x - 2)^2 + (y + 1)^2 = 16$?
A) $(2, -1)$ B) $(1, -2)$ C) $(3, 1)$ D) $(4, 2)$
Correct Answer: A Explanation: The center of a circle with the equation $(x - 2)^2 + (y + 1)^2 = 16$ is $(2, -1)$.
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