College Algebra, Chapter 2, Review Exercises, Section Review Exercises, Problem 4

Suppose that $P(5,-2)$ and $Q(-3,-6)$
a.) Plot $P$ and $Q$ on a coordinate plane.



b.) Find the distance from $P$ to $Q$.
By using distance formula

$
\begin{equation}
\begin{aligned}
d &= \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}\\
\\
d &= \sqrt{(-6-(-2))^2 + (-3-5)^2}\\
\\
d &= \sqrt{(-6+2)^2 + (-3-5)^2}\\
\\
d &= \sqrt{(-4)^2 + (-8)^2}\\
\\
d &= \sqrt{80} = 4\sqrt{5} \text{ units}
\end{aligned}
\end{equation}
$


c.) Find the midpoint of the segment $PQ$.

$
\begin{equation}
\begin{aligned}
m_{PQ}, \quad x &= \frac{5-3}{2} = 1\\
\\
y &= \frac{-2-6}{2} = -4
\end{aligned}
\end{equation}
$

Therefore, $m_{PQ}(1,-4)$

d.) Find the slope of the line determined by $P$ and $Q$ and find equations for the line in point slope form and in slope intercept form. Then sketch a graph of the line.
$\displaystyle m = \frac{y_2-y_1}{x_2-x_1} = \frac{-6-(-2)}{-3-5} = \frac{-6+2}{-8} = \frac{-4}{-8} = \frac{1}{2}$
Thus, by using point slope form

$
\begin{equation}
\begin{aligned}
y- y_1 &= m(x-x_1) \\
\\
y - (-2) &= \frac{1}{2}x (x-5)\\
\\
y +2 &= \frac{1}{2} x + \frac{5}{2}\\
\\
y &= \frac{1}{2} x - \frac{9}{2}
\end{aligned}
\end{equation}
$

By using slope intercept form,

$
\begin{equation}
\begin{aligned}
y &= mx + b\\
\\
y &= \frac{1}{2} x + b
\end{aligned}
\end{equation}
$


@ $P(5,-2)$

$
\begin{equation}
\begin{aligned}
-2 &= \frac{1}{2} (5) + b && \text{Solve for } b\\
\\
b &= -2 -\frac{5}{2} = \frac{-9}{2}\\
\\
\text{Thus,}\\
\\
y &= \frac{1}{2} x - \frac{9}{2}
\end{aligned}
\end{equation}
$






e.) Sketch the circle that passes through $Q$ and has center $P$, and find the equation of that circle.
Recall that the general equation of the circle with center $(h,k)$ and radius $r$ is...
$(x-h)^2 + (y-k)^2 = r^2$
Since the center is at $P(5,-2)$
$(x-5)^2 + (y+2)^2 = r^2$
And its pass through $Q(-3,-6)$
$(-3-5)^2 + (-6+2) = r^2$
$(-8)^2 + (-4)^2 = r^2$

Solving for $r^2$,
$r = 80$

Thus, the equation of the circle is...
$(x - 5)^2 + (y + 2) = 80$