Problem 48 - Precalculus

Question

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. $$3 x^{2}+3 y^{2}+5 x-4 y=1$$

Step-by-Step Solution

Answer

Center: \((-\frac{5}{6}, \frac{2}{3})\), Radius: \(\frac{\sqrt{53}}{6}\), y-intercepts: \(\frac{2}{3} \pm \frac{\sqrt{7}}{3}\).

Step 1: 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.

Step 2: 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.

Step 3: 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:\[ (x + \frac{5}{6})^2 - \frac{25}{36} \].

Step 4: 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:\[ (y - \frac{2}{3})^2 - \frac{4}{9} \].

Step 5: 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.

Step 6: 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:\[ \frac{12}{36} + \frac{25}{36} + \frac{16}{36} = \frac{53}{36} \].

Step 7: 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 \(\sqrt{\frac{53}{36}} = \frac{\sqrt{53}}{6}\).

Step 8: 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:\[ y - \frac{2}{3} = \pm \sqrt{\frac{7}{9}} = \pm \frac{eq{\sqrt{7}}}{3} \].\[ y = \frac{2}{3} \pm \frac{eq{\sqrt{7}}}{3} \].

Key Concepts

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 \((-\frac{5}{6}, \frac{2}{3})\).

This equation showcases how manipulating the equation through algebraic techniques, such as completing the square, helps identify the exact center of the circle.

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 \[\frac{\sqrt{53}}{6}\].

Understanding the calculation of the radius helps in visualizing the size of the circle in terms of the units on a graph or a coordinate plane.

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 \[(y - \frac{2}{3})^2 - \frac{4}{9}\].

These transformations simplify the quadratic into parts where the center coordinates and the enabling distances (radius) surface cleanly, making the geometric interpretation straightforward.

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 \(y = \frac{2}{3} \pm \frac{\sqrt{7}}{3}\).

Intersections such as these assist greatly in constructing a graph and identifying symmetry or key features relevant to problem-solving and analysis in geometry.

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—center, radius, and intersections—via 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.

Ultimately, mastering coordinate geometry techniques enables deeper insights into not just circles, but a variety of geometric entities and their applications.