Tuesday, January 14, 2020

College Algebra, Chapter 2, 2.3, Section 2.3, Problem 28

Graph the circle $(y-1)^2 + x^2 =1$ by solving for $y$ and graphing two equations involved.

$
\begin{equation}
\begin{aligned}
(y-1)^2 + x^2 &= 1 && \text{Model}\\
\\
(y-1)^2 &= 1-x^2 && \text{Subtract } x^2\\
\\
y - 1 &= \pm \sqrt{1-x^2} && \text{Take the square root}\\
\\
y &= 1 \pm \sqrt{1-x^2} && \text{Add } 1\\
\\
y &= 1 + \sqrt{1 - x^2} \text{ and } y = 1- \sqrt{1-x^2} && \text{Solve for }y
\end{aligned}
\end{equation}
$

Thus, the circle is descended by the graphs of the equations
$y =1 + \sqrt{1+x^2}$ and $y = 1 - \sqrt{1 - x^2}$

The first equation represents the top half of the circle because $y \geq 0$ while the second represents the bottom half. If we graph the first equation in the viewing rectangle $[-1,1]$ by $[-2,2]$ then we get...


The graph of the second equation is....



Graphing the semicircles together on the same viewing screen, we get the full circle...

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