Chapter 2: Problem 38
Find the center \((h, k)\) and radius \(r\) of each circle; \((b)\) graph each circle; \((c)\) find the intercepts, if any. $$ 3 x^{2}+3 y^{2}-12 y=0 $$
Short Answer
Expert verified
Center: \((0, 2)\), Radius: 2, Intercepts: \((0,0)\), \((0, 4)\)
Step by step solution
01
- Divide by the coefficient of squared terms
Given the equation: \[ 3x^2 + 3y^2 - 12y = 0 \]Divide every term by 3 to simplify: \[ x^2 + y^2 - 4y = 0 \]
02
- Complete the square
To complete the square for the y-terms, rewrite the equation as: \[ x^2 + (y^2 - 4y) = 0 \]Take half of the coefficient of y, square it, and add and subtract it inside the equation:\[ x^2 + (y^2 - 4y + 4 - 4) = 0 \]This becomes: \[ x^2 + (y - 2)^2 - 4 = 0 \]Add 4 to both sides to isolate the perfect square: \[ x^2 + (y-2)^2 = 4 \]
03
- Identify the circle's center and radius
The standard equation of a circle is \[ (x - h)^2 + (y - k)^2 = r^2 \]Comparing this with \[ x^2 + (y - 2)^2 = 4 \], we see that the center is \((h, k) = (0, 2)\) and the radius is \(r\) such that \(r^2 = 4\), hence \(r = 2\).
04
- Graph the circle
To graph the circle, plot the center at \((0, 2)\), then use the radius of 2 units to draw a circle around the center, extending 2 units in all directions (up, down, left, right).
05
- Find the intercepts
To find the intercepts, first find the x-intercepts by setting \(y = 0\) and solving for \(x\):\[ x^2 + (0 - 2)^2 = 4 \]\therefore:\[ x^2 + 4 = 4 \]\[ x^2 = 0 \]\[ x = 0 \]So the x-intercept is \((0, 0)\). To find the y-intercepts, set \(x = 0\):\[ (0)^2 + (y - 2)^2 = 4 \]\therefore:\[ (y - 2)^2 = 4 \]\[ y - 2 = ±2 \]\[ y = 4 \ or\ y = 0 \]So, the y-intercepts are \((0, 4)\) and \((0, 0)\).
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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 crucial technique for simplifying quadratic equations, especially when dealing with circles. The idea is to transform a quadratic expression into a perfect square trinomial. This makes it easier to identify the center and radius of a circle.
To complete the square for an equation like \( x^2 + y^2 - 4y = 0 \), follow these steps:
\ x^2 + (y - 2)^2 - 4 = 0 \.
By isolating the perfect square, you make it easy to convert the equation into the standard form of a circle.
To complete the square for an equation like \( x^2 + y^2 - 4y = 0 \), follow these steps:
- Focus on the terms involving \( y \). Here, it's \( y^2 - 4y \).
- Take half of the coefficient of \( y \), which is \( -4 \), divide by 2 to get \( -2 \), and then square it to get \( 4 \).
- Add and subtract this square inside the equation, turning \( y^2 - 4y \) into \( (y - 2)^2 - 4 \).
\ x^2 + (y - 2)^2 - 4 = 0 \.
By isolating the perfect square, you make it easy to convert the equation into the standard form of a circle.
center and radius of a circle
Understanding the center and radius of a circle is critical. The standard form of a circle equation is:
\ (x - h)^2 + (y - k)^2 = r^2 \.
Here, \( (h, k) \) represents the center, and \( r \) represents the radius.
From our simplified equation \ x^2 + (y - 2)^2 = 4 \, we can identify:
\ (x - h)^2 + (y - k)^2 = r^2 \.
Here, \( (h, k) \) represents the center, and \( r \) represents the radius.
From our simplified equation \ x^2 + (y - 2)^2 = 4 \, we can identify:
- Center \( (h, k) = (0, 2) \)
- Radius \( r \) as \( 2 \, because \ r^2 = 4 \).
intercepts of a circle
Finding intercepts involves determining where the circle crosses the x-axis and y-axis. To find intercepts, set the opposite variable to zero and solve the resulting equation.
For x-intercepts, set \ y = 0 \: \ x^2 + (0 - 2)^2 = 4 \
Simplify to find:
\ x^2 + 4 = 4 \ \Rightarrow x^2 = 0 \ \Rightarrow x = 0 \
So, the x-intercept is \( (0, 0) \).
For y-intercepts, set \ x = 0 \: \ (0)^2 + (y - 2)^2 = 4 \
This simplifies to:
\ (y - 2)^2 = 4 \ \Rightarrow y - 2 = \pm 2 \Rightarrow y = 0 \ or \ 4 \
Thus, the y-intercepts are \( (0, 0) \) and \( (0, 4) \).
Understanding intercepts is essential for graphing and analyzing the behavior of a circle on a coordinate plane.
For x-intercepts, set \ y = 0 \: \ x^2 + (0 - 2)^2 = 4 \
Simplify to find:
\ x^2 + 4 = 4 \ \Rightarrow x^2 = 0 \ \Rightarrow x = 0 \
So, the x-intercept is \( (0, 0) \).
For y-intercepts, set \ x = 0 \: \ (0)^2 + (y - 2)^2 = 4 \
This simplifies to:
\ (y - 2)^2 = 4 \ \Rightarrow y - 2 = \pm 2 \Rightarrow y = 0 \ or \ 4 \
Thus, the y-intercepts are \( (0, 0) \) and \( (0, 4) \).
Understanding intercepts is essential for graphing and analyzing the behavior of a circle on a coordinate plane.