# Distance, Circles, and Semi-circles

1. Find the distance between each pair of points given:

1. $(-1,3)$, $(2,4)$.   $\ans{\sqrt{(4-3)^2 + (2 - (-1))^2} = \sqrt{10}}$

2. $(-3,2)$, $(1,-7)$.   $\ans{\sqrt{(-7-2)^2 + (1-(-3))^2} = \sqrt{97}}$

2. Find the midpoint for each pair of points given:

1. $(-4,1), (2,-5)$   $\ans{\displaystyle{\left( \frac{(-4)+2}{2}, \frac{1+(-5)}{2} \right) = (-1,-2)}}$

2. $(-1,3), (2,4)$   $\ans{\displaystyle{\left( \frac{(-1) + 2}{2}, \frac{3+4}{2} \right) = (\frac{1}{2}, \frac{7}{2})}}$

3. $(-3,2)$, $(1,-7)$   $\ans{\displaystyle{\left( \frac{(-3) + 1}{2}, \frac{2 + (-7)}{2} \right) = (-1,-\frac{5}{2})}}$

3. Find the center and radius for each of the following:

1. $x^2 + y^2 - 6x + 10y + 9 = 0$   $\ans{ \begin{array}{l} \textrm{To find center and radius we put the equation in the }\\ \textrm{standard form for a circle by completing the square in}\\ \textrm{both } x \textrm{ and } y:\\\\ \begin{array}{rcl} x^2+y^2-6x+10y+9 &=& 0\\ (x^2 - 6x + 9) + (y^2 + 10y + 25) &=& 25\\ (x-3)^2 + (y+5)^2 &=& 25 \end{array}\\\\ \textrm{center: } (3,-5)\\ \textrm{radius: }\sqrt{25} = 5 \end{array}}$

2. $x^2 + y^2 + 6x + 12y - 5 = 0$   $\ans{ \begin{array}{l} \textrm{To find center and radius we put the equation in the }\\ \textrm{standard form for a circle by completing the square in}\\ \textrm{both } x \textrm{ and } y:\\\\ \begin{array}{rcl} x^2 + y^2 + 6x + 12y - 5 &=& 0\\ (x^2 + 6x + 9) + (y^2 + 12y + 36) &=& 5 + 9 + 36\\ (x+3)^2 + (y+6)^2 &=& 50 \end{array}\\\\ \textrm{center: } (-3,-6)\\ \textrm{radius: }\sqrt{50} = 5\sqrt{2} \end{array}}$

4. Find the equation of the circle with center $(-1,5)$ and radius of $6$.   $\ans{\displaystyle{(x+1)^2 + (y-5)^2 = 36}}$

5. Find the equation for the circle with endpoints of a diameter at $(-2,0)$ and $(4,8)$.   $\ans{(x-1)^2 + (y-4)^2 = 25}$

6. Sketch the graph of $x^2+y^2+8x-6y+16=0$, labeling all important aspects.