Circles

1. Proof:

Let $\odot O_1$ have center $O_1$ and radius $r_1$, and $\odot O_2$ have center $O_2$ and radius $r_2$.

Step 1 — Translation: Translate $\odot O_1$ by the vector $\overrightarrow{O_1 O_2}$. This maps $O_1$ to $O_2$, producing a circle centered at $O_2$ with radius $r_1$. Translations preserve size and shape.

Step 2 — Dilation: Apply a dilation centered at $O_2$ with scale factor $k = \dfrac{r_2}{r_1}$. This maps the translated circle (radius $r_1$, center $O_2$) to a circle with radius $\dfrac{r_2}{r_1} \cdot r_1 = r_2$, centered at $O_2$.

The result is exactly $\odot O_2$. Since a translation followed by a dilation is a similarity transformation, the two circles are similar.

Since $O_1$, $r_1$, $O_2$, and $r_2$ were arbitrary, all circles are similar. $\blacksquare$

2. a) The Inscribed Angle Theorem states that an inscribed angle is half the central angle that intercepts the same arc:

$\angle ACB = \frac{1}{2} \cdot \angle AOB = \frac{1}{2} \cdot 110° = 55°$

b) $\angle ADB = 55°$. All inscribed angles that intercept the same arc are congruent. Since $\angle ADB$ intercepts the same arc $\overset{\frown}{AB}$ as $\angle ACB$, they are equal.

c) Point $C$ is on the major arc (the arc that does not contain the central angle's interior), so $\angle ACB$ intercepts the minor arc $\overset{\frown}{AB} = 110°$.

Point $E$ is on the minor arc (opposite side from $C$). An inscribed angle intercepts the arc on the opposite side from the vertex, so $\angle AEB$ intercepts the major arc $\overset{\frown}{AB} = 360° - 110° = 250°$:

$\angle AEB = \frac{1}{2} \cdot 250° = 125°$

Note that $\angle ACB + \angle AEB = 55° + 125° = 180°$. This always holds when two inscribed angles intercept complementary arcs (minor and major) of the same chord.

3. a) The arc length formula is $s = \frac{\theta}{360°} \cdot 2\pi r$:

$s = \frac{72°}{360°} \cdot 2\pi(10) = \frac{1}{5} \cdot 20\pi = 4\pi \approx 12.57 \text{ cm}$

b) The sector area formula is $A = \frac{\theta}{360°} \cdot \pi r^2$:

$A = \frac{72°}{360°} \cdot \pi(10)^2 = \frac{1}{5} \cdot 100\pi = 20\pi \approx 62.83 \text{ cm}^2$

c) $\frac{72°}{360°} = \frac{1}{5}$. The sector represents $\dfrac{1}{5}$ (or $20\%$) of the full circle.

4. The points appear in the order $A, C, B, D$ around the circle. When two chords intersect inside a circle, the measure of an angle formed equals half the sum of the two arcs it intercepts (the arc between one pair of endpoints and the arc between the other pair).

a) $\angle APC$ is formed by chords $\overline{AB}$ and $\overline{CD}$. It intercepts arcs $\overset{\frown}{AC}$ and $\overset{\frown}{BD}$ (the arcs "across" from the angle):

$\angle APC = \frac{1}{2}(\overset{\frown}{AC} + \overset{\frown}{BD}) = \frac{1}{2}(80° + 40°) = 60°$

b) The four arcs in order around the circle are $\overset{\frown}{AC} = 80°$, $\overset{\frown}{CB}$, $\overset{\frown}{BD} = 40°$, $\overset{\frown}{DA} = 140°$:

$80° + \overset{\frown}{CB} + 40° + 140° = 360°$

$\overset{\frown}{CB} = 360° - 260° = 100°$

c) Verify: $80° + 100° + 40° + 140° = 360°$ $\checkmark$

Now find $\angle APD$. This angle is supplementary to $\angle APC$ (they form a linear pair along chord $\overline{CD}$), and it intercepts arcs $\overset{\frown}{DA}$ and $\overset{\frown}{CB}$:

$\angle APD = \frac{1}{2}(\overset{\frown}{DA} + \overset{\frown}{CB}) = \frac{1}{2}(140° + 100°) = 120°$

Confirm: $\angle APC + \angle APD = 60° + 120° = 180°$ (supplementary, as expected for a linear pair formed by intersecting chords). $\checkmark$

5. The radius is $r = \frac{16}{2} = 8$ inches and the central angle is $45°$.

a) Sector area:

$A = \frac{45°}{360°} \cdot \pi r^2 = \frac{1}{8} \cdot \pi(8)^2 = \frac{1}{8} \cdot 64\pi = 8\pi \approx 25.13 \text{ in}^2$

b) Arc length:

$s = \frac{45°}{360°} \cdot 2\pi r = \frac{1}{8} \cdot 2\pi(8) = \frac{1}{8} \cdot 16\pi = 2\pi \approx 6.28 \text{ in}$

c) The slice has two straight edges (radii of length $8$ in) and one curved edge:

$\text{Perimeter} = 2r + s = 2(8) + 2\pi = 16 + 2\pi \approx 22.28 \text{ in}$