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Magazine Chapter 1: Problem 48
Determine the center and the radius for the circle. Also, find the \(y\) -coordinates of the points (if any) where the circle intersects the \(y\) -axis. $$3 x^{2}+3 y^{2}+5 x-4 y=1$$
Short Answer
Expert verified
Center: \((-\frac{5}{6}, \frac{2}{3})\), Radius: \(\frac{\sqrt{53}}{6}\), y-intercepts: \(\frac{2}{3} \pm \frac{\sqrt{7}}{3}\).
Step by step solution
01
Rewrite the Circle Equation
First, divide the entire equation by 3 to simplify it: \[ x^2 + y^2 + \frac{5}{3}x - \frac{4}{3}y = \frac{1}{3} \]. This is done to make the coefficients of \(x^2\) and \(y^2\) equal to 1.
02
Group Similar Terms
Group the terms related to \(x\) and \(y\): \[ (x^2 + \frac{5}{3}x) + (y^2 - \frac{4}{3}y) = \frac{1}{3} \]. We will complete the square for each of these groups separately.
03
Complete the Square for x
To complete the square for the \(x\) terms, use the formula \((x + \frac{b}{2a})^2\) where \(b = \frac{5}{3}\). Add and subtract \(\left(\frac{5}{6}\right)^2 = \frac{25}{36}\) inside the equation:\[ (x + \frac{5}{6})^2 - \frac{25}{36} \].
04
Complete the Square for y
For the \(y\) terms, use the formula \((y + \frac{b}{2a})^2\) where \(b = \frac{-4}{3}\). Add and subtract \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\) inside the equation:\[ (y - \frac{2}{3})^2 - \frac{4}{9} \].
05
Combine Completed Squares
Combine the completed squares in the equation: \[ (x + \frac{5}{6})^2 + (y - \frac{2}{3})^2 = \frac{1}{3} + \frac{25}{36} + \frac{4}{9} \]. Simplify the right-hand side.
06
Simplify Right-hand Side
Calculate the right-hand side of the equation:Convert fractions to have a common denominator, the least common multiple being 36:\[ \frac{1}{3} = \frac{12}{36}, \quad \frac{4}{9} = \frac{16}{36} \].Adding them up:\[ \frac{12}{36} + \frac{25}{36} + \frac{16}{36} = \frac{53}{36} \].
07
Find Circle's Center and Radius
Rewrite the circle's equation:\[ (x+\frac{5}{6})^2 + (y-\frac{2}{3})^2 = \frac{53}{36} \].The center is \((-\frac{5}{6}, \frac{2}{3})\) and the radius is \(\sqrt{\frac{53}{36}} = \frac{\sqrt{53}}{6}\).
08
Find y-intercepts
To find where the circle intersects the \(y\)-axis, set \(x = 0\) in the equation:\[ (0 + \frac{5}{6})^2 + (y - \frac{2}{3})^2 = \frac{53}{36} \].Solve for \(y\):\[ (\frac{5}{6})^2 + (y - \frac{2}{3})^2 = \frac{53}{36} \].\[ \frac{25}{36} + (y - \frac{2}{3})^2 = \frac{53}{36} \].\[ (y - \frac{2}{3})^2 = \frac{28}{36} = \frac{7}{9} \].Solving gives:\[ y - \frac{2}{3} = \pm \sqrt{\frac{7}{9}} = \pm \frac{eq{\sqrt{7}}}{3} \].\[ y = \frac{2}{3} \pm \frac{eq{\sqrt{7}}}{3} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Center of Circle
The center of a circle in the coordinate plane provides crucial information about its position and orientation. To determine the center from the equation of a circle, we rewrite the equation in the standard form:
- The circle equation is given by \[(x - h)^2 + (y - k)^2 = r^2\]where \((h, k)\) is the center, and \(r\) is the radius of the circle.
- In our exercise, after completing the square method, the circle's equation becomes \[(x + \frac{5}{6})^2 + (y - \frac{2}{3})^2 = \frac{53}{36}\]. Thus, the center is \((-\frac{5}{6}, \frac{2}{3})\).
Radius of Circle
The radius of a circle is the distance from its center to any point on the circle itself.
- From the standard circle equation \[(x - h)^2 + (y - k)^2 = r^2\], the term \(r^2\) represents the square of the radius.
- Our task involves finding this distance from the given equation. After simplifying, the equation reads \[(x + \frac{5}{6})^2 + (y - \frac{2}{3})^2 = \frac{53}{36}\].We discover that \(r^2 = \frac{53}{36}\), so the radius \(r\) is \[\frac{\sqrt{53}}{6}\].
Completing the Square
Completing the square is a method used in algebra to transform a quadratic equation into a form that reveals the maximum, minimum, or other notable features, such as the circle's center and radius in coordinate geometry.
- The process involves making modifications to the equation to create perfect square terms.
- For the quadratic terms in\(x\) and\(y\), we adjust them respectively as follows: \[(x^2 + \frac{5}{3}x) \]becomes \[(x + \frac{5}{6})^2 - \frac{25}{36}\], and\[(y^2 - \frac{4}{3}y) \]becomes \[(y - \frac{2}{3})^2 - \frac{4}{9}\].
Circle Intersections
Finding where a circle intersects the axes, particularly the \(y\)-axis, provides valuable points that contribute to understanding the circle's position relative to standard reference lines.
- For instance, setting \(x = 0\) in the circle equation helps in finding points where the circle intersects the \(y\)-axis.
- In the current example, substituting \(x = 0\) yields the expression: \[(0 + \frac{5}{6})^2 + (y - \frac{2}{3})^2 = \frac{53}{36}\],leading to finding\[y - \frac{2}{3} = \pm \frac{\sqrt{7}}{3}\].This results in two potential \(y\)-intercepts at \(y = \frac{2}{3} \pm \frac{\sqrt{7}}{3}\).
Coordinate Geometry
Coordinate geometry, or analytic geometry, bridges algebra with geometry through graphs and numerical coordinates for representing geometric shapes. This linkage enhances understanding of shapes like circles by using a coordinate system.
- In the context of a circle, coordinate geometry allows for precise descriptions of its features鈥攃enter, radius, and intersections鈥攙ia algebraic expressions.
- Through converting and manipulating equations, as done via completing the square, students can determine where exactly a circle sits on a plane, its size, and its interaction with axes.
