Answer
$(a)$ See the image below.
$(b)$
$d=\sqrt{89}$
$(c)$
$M_{PQ}(1, 3.5)$
$(d)$
$m=\frac{5}{8}$
$(e)$
$y=-1.6x+5.1$
$(f)$
$(x – 1)^2 + (y – 3.5)^2 = 22.25$
Work Step by Step
$(a)$ See the image above.
$(b)$ We can find the distance using distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$d=\sqrt{(5+3)^2+(6-1)^2}=\sqrt{64+25}=\sqrt{89}$
$d=\sqrt{89}$
$(c)$ We can locate the midpoint using midpoint formula: $M_{PQ}=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$
$M_{PQ}=(\frac{-3+5}{2}, \frac{1+6}{2})=(1, 3.5)$
$(d)$ Slope $m$ of a line is calculated by dividing change in $y$ by change in $x$ for any given two points of the line.
$m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-1}{5+3}=\frac{5}{8}$
$m=\frac{5}{8}$
$(e)$ Using the perpendicular bisector form of a line $(x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2$
We can write:
$(x+3)^2+(y-1)^2=(x-5)^2+(y-6)^2$
$x^2+6x+9+y^2-2y+1=x^2-10x+25+y^2-12y+36$
$10y=-16x+51$
$y=-1.6x+5.1$
$(f)$ To find the equation of the circle, we will use the following form of a circle:
$(x – a)^2 + (y – b)^2 = r^2$
Where $r$ is a radius of the circle and $a$ and $b$ are coordinates of the center of the circle; in our case, the coordinates of $M$.
We also know the diameter of the circle (The distance between $P$ and $Q$), so the radius is its half.
$r=\frac{\sqrt{89}}{2}$
$(x – 1)^2 + (y – 3.5)^2 = (\frac{\sqrt{89}}{2})^2$
$(x – 1)^2 + (y – 3.5)^2 = \frac{\sqrt{89}}{4}$
$(x – 1)^2 + (y – 3.5)^2 = 22.25$