Coordinate Geometry

1. The distance from any point $(x, y)$ on the circle to the center $(3, -2)$ equals the radius $5$:

$\sqrt{(x - 3)^2 + (y + 2)^2} = 5$

Squaring both sides:

$(x - 3)^2 + (y + 2)^2 = 25$

To check the point $(6, 2)$, substitute:

$(6 - 3)^2 + (2 + 2)^2 = 9 + 16 = 25$

Since $25 = 25$ (equals $r^2$), the point $(6, 2)$ lies on the circle.

2. a) The center lies on $y = 2$, so the center has coordinates $(h, 2)$. Since $A(1, 5)$ and $B(7, 5)$ both lie on the circle, the center is equidistant from both. Because $A$ and $B$ have the same $y$-coordinate, the center lies on the perpendicular bisector of $\overline{AB}$, which is the vertical line $x = \frac{1 + 7}{2} = 4$.

Center: $(4, 2)$.

b) The radius is the distance from the center to either point:

$r = \sqrt{(4 - 1)^2 + (2 - 5)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$

c) $(x - 4)^2 + (y - 2)^2 = 18$

Verification: Check point $B(7, 5)$: $(7-4)^2 + (5-2)^2 = 9 + 9 = 18$ $\checkmark$

3. Proving it is a parallelogram — show that opposite sides are parallel (equal slopes):

Slope of $\overline{AB}$: $\frac{0 - 0}{5 - 0} = 0$

Slope of $\overline{DC}$: $\frac{4 - 4}{7 - 2} = 0$

Since slope of $\overline{AB}$ = slope of $\overline{DC}$, $\overline{AB} \parallel \overline{DC}$.

Slope of $\overline{AD}$: $\frac{4 - 0}{2 - 0} = 2$

Slope of $\overline{BC}$: $\frac{4 - 0}{7 - 5} = 2$

Since slope of $\overline{AD}$ = slope of $\overline{BC}$, $\overline{AD} \parallel \overline{BC}$.

Both pairs of opposite sides are parallel, so $ABCD$ is a parallelogram. $\checkmark$

Checking for rectangle — a parallelogram is a rectangle if and only if its diagonals are equal in length:

$AC = \sqrt{(7 - 0)^2 + (4 - 0)^2} = \sqrt{49 + 16} = \sqrt{65}$

$BD = \sqrt{(2 - 5)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

Since $\sqrt{65} \neq 5$, the diagonals are not equal. Therefore $ABCD$ is not a rectangle.

(Alternatively, $\overline{AB}$ has slope $0$ and $\overline{AD}$ has slope $2$. Since $0 \times 2 \neq -1$, the sides are not perpendicular, confirming it is not a rectangle.)

4. a) Compute each side length:

$PQ = \sqrt{(7-1)^2 + (1-1)^2} = \sqrt{36} = 6$

$QR = \sqrt{(4-7)^2 + (9-1)^2} = \sqrt{9 + 64} = \sqrt{73}$

$PR = \sqrt{(4-1)^2 + (9-1)^2} = \sqrt{9 + 64} = \sqrt{73}$

Perimeter:

$P = 6 + \sqrt{73} + \sqrt{73} = 6 + 2\sqrt{73}$

b) Midpoint of $\overline{QR}$:

$M = \left(\frac{7 + 4}{2},\; \frac{1 + 9}{2}\right) = \left(\frac{11}{2},\; 5\right)$

c) The median from $P$ to $M$:

$PM = \sqrt{\left(\frac{11}{2} - 1\right)^2 + (5 - 1)^2} = \sqrt{\left(\frac{9}{2}\right)^2 + 4^2} = \sqrt{\frac{81}{4} + 16} = \sqrt{\frac{81 + 64}{4}} = \sqrt{\frac{145}{4}} = \frac{\sqrt{145}}{2}$

And $\frac{1}{2} QR = \frac{\sqrt{73}}{2}$.

Since $\frac{\sqrt{145}}{2} \neq \frac{\sqrt{73}}{2}$ (i.e., $145 \neq 73$), $PM \neq \frac{1}{2} QR$.

Note: Since $PR = QR = \sqrt{73}$, triangle $PQR$ is actually isosceles with $PR = QR$. The median from $P$ to the midpoint of $QR$ has length $\frac{\sqrt{145}}{2} \approx 6.02$.

5. a) The center is the midpoint of the diameter:

$\text{Center} = \left(\frac{-3+5}{2},\; \frac{1+7}{2}\right) = (1, 4)$

The radius is half the diameter length:

$d = \sqrt{(5-(-3))^2 + (7-1)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$

$r = \frac{10}{2} = 5$

b) $(x - 1)^2 + (y - 4)^2 = 25$

c) Point $C(5, 1)$ lies on the circle:

$(5 - 1)^2 + (1 - 4)^2 = 16 + 9 = 25 = r^2 \quad \checkmark$

Proving $\angle ACB = 90°$:

Compute the slopes of $\overline{CA}$ and $\overline{CB}$:

$m_{CA} = \frac{1 - 1}{-3 - 5} = \frac{0}{-8} = 0$

$m_{CB} = \frac{7 - 1}{5 - 5} = \frac{6}{0} \implies \text{undefined (vertical line)}$

$\overline{CA}$ is horizontal (slope $0$) and $\overline{CB}$ is vertical (undefined slope). A horizontal and a vertical line are perpendicular.

Therefore $\angle ACB = 90°$, confirming Thales' theorem. $\blacksquare$