Chapter 1: Problem 103
Show that the equation represents a circle, and find the center and radius of the circle. $$2 x^{2}+2 y^{2}-3 x=0$$
Short Answer
Expert verified
This equation represents a circle with center \(\left(\frac{3}{4}, 0\right)\) and radius \(\frac{3}{4}\).
Step by step solution
01
Rewrite the Equation in Standard Form
The given equation is \( 2x^2 + 2y^2 - 3x = 0 \). Start by dividing every term by 2 to simplify the equation:\[ x^2 + y^2 - \frac{3}{2}x = 0 \]
02
Rearrange Terms for Completing the Square
Rearrange the equation by isolating the terms involving \(x\) and \(y\):\[ x^2 - \frac{3}{2}x + y^2 = 0 \]
03
Complete the Square for the x-Term
To complete the square for the \(x\) terms, take the coefficient of \(x\) which is \(-\frac{3}{2}\), divide by 2, and square it. This gives:\[ \left(-\frac{3}{4}\right)^2 = \frac{9}{16} \]Add and subtract \(\frac{9}{16}\) inside the equation for the \(x\) terms:\[ \left( x^2 - \frac{3}{2}x + \frac{9}{16} \right) - \frac{9}{16} + y^2 = 0 \]
04
Express x-Terms as a Perfect Square
Rewrite the completed square for \(x\) to simplify the expression:\[ \left(x - \frac{3}{4}\right)^2 + y^2 = \frac{9}{16} \]
05
Express the Equation in Standard Circle Form
The equation \(\left(x - \frac{3}{4}\right)^2 + y^2 = \frac{9}{16} \) is now in the standard form of a circle equation, which is \((x - h)^2 + (y - k)^2 = r^2\). Here, \(h = \frac{3}{4}\), \(k = 0\), and \(r^2 = \frac{9}{16}\).
06
Determine the Center and Radius of the Circle
The center \((h, k)\) of the circle is \(\left(\frac{3}{4}, 0\right)\), and the radius \(r\) is the square root of \(\frac{9}{16}\), which is \(\frac{3}{4}\).
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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 equation into a perfect square trinomial. This approach makes it easier to graph ellipses and circles as it allows us to rewrite the equation in a standardized form. Consider the equation from our exercise:
\[ x^2 - \frac{3}{2}x + y^2 = 0 \]To complete the square for the \( x \) terms, you identify the linear coefficient, which is \(-\frac{3}{2}\) in this case. Divide it by 2, resulting in \(-\frac{3}{4}\). Squaring \(-\frac{3}{4}\) yields \( \frac{9}{16} \).
You then add and subtract this squared value inside the expression:
\[ \left( x - \frac{3}{4} \right)^2 \]
Completing the square transforms parts of the equation into a form that reveals geometrical properties more clearly. Next, we can find the center of the circle.
\[ x^2 - \frac{3}{2}x + y^2 = 0 \]To complete the square for the \( x \) terms, you identify the linear coefficient, which is \(-\frac{3}{2}\) in this case. Divide it by 2, resulting in \(-\frac{3}{4}\). Squaring \(-\frac{3}{4}\) yields \( \frac{9}{16} \).
You then add and subtract this squared value inside the expression:
- Add \( \frac{9}{16} \) to complete the square.
- Subtract \( \frac{9}{16} \) to maintain the balance of the equation.
\[ \left( x - \frac{3}{4} \right)^2 \]
Completing the square transforms parts of the equation into a form that reveals geometrical properties more clearly. Next, we can find the center of the circle.
Center of a Circle
The center of a circle in its standard equation form \((x - h)^2 + (y - k)^2 = r^2\)is denoted by the point \((h, k)\). In geometrical terms, this point is equidistant from all points along the circle's boundary.
From the equation \((x - \frac{3}{4})^2 + y^2 = \frac{9}{16}\), we can identify \(h = \frac{3}{4}\) and \(k = 0\). This tells us that:
From the equation \((x - \frac{3}{4})^2 + y^2 = \frac{9}{16}\), we can identify \(h = \frac{3}{4}\) and \(k = 0\). This tells us that:
- The center is located at the point \(\left(\frac{3}{4}, 0\right)\).
- This position means the circle is shifted horizontally by \(\frac{3}{4}\) units from the origin along the x-axis.
Radius of a Circle
The radius of a circle is defined as the distance from the center of the circle to any point on its circumference. This measurement is a constant for any circle and is denoted by \( r \).
In the standard circle formula \((x - h)^2 + (y - k)^2 = r^2\), \(r^2\) is explicitly given. For our example:
In the standard circle formula \((x - h)^2 + (y - k)^2 = r^2\), \(r^2\) is explicitly given. For our example:
- \(r^2 = \frac{9}{16}\)
- To find the radius \(r\), take the square root of both sides, resulting in \(r = \frac{3}{4}\).