Notes
Parts of a circle
Radius, diameter ($d = 2r$), chord, secant, tangent. Circumference $C = 2\pi r$.
Angle and arc rules
| Angle | Measure |
|---|---|
| Central angle | Equal to its arc |
| Inscribed angle | Half its arc |
| Tangent and chord | Half its arc |
| Two chords crossing inside | Half the sum of the arcs |
| Vertex outside the circle | Half the difference of the arcs |
- An angle inscribed in a semicircle is $90^\circ$. Opposite angles of an inscribed quadrilateral are supplementary.
Segment rules
- A tangent is perpendicular to the radius at the point of tangency.
- Two tangent segments from one outside point are congruent.
- Crossing chords: products of the pieces are equal.
- A diameter perpendicular to a chord bisects it.
Measurement and equations
- Arc length $\displaystyle = \frac{\theta}{360^\circ} \cdot 2\pi r$. Sector area $\displaystyle = \frac{\theta}{360^\circ} \cdot \pi r^2$. Radians: $180^\circ = \pi$.
- Circle: $(x - h)^2 + (y - k)^2 = r^2$. Complete the square to find the center and radius.
Topics in this unit
Each topic has its own short notes and practice.
Practice
The arc is $96^\circ$, and a central angle equals its arc.
The distance from $P$ to the center is $\displaystyle \sqrt{7^2 + 24^2} = 25$. Subtract the radius: $25 - 7 = 18$.
$\displaystyle r = \sqrt{3^2 + 4^2} = 5$, so $r^2 = 25$.
$\displaystyle \frac{250 - 110}{2} = 70$.
Radius 8: total area $64\pi$, and $\displaystyle \frac{64\pi}{8} = 8\pi$.
$\displaystyle \frac{72}{360} \cdot 20\pi = \frac{1}{5} \cdot 20\pi = 4\pi$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A lawn sprinkler sits at point $S(4, -3)$ on a yard map measured in feet. It sprays water 15 feet in every direction it points.
(a) The sprinkler turns through $120^\circ$. Find the area it waters, exactly and to the nearest tenth.
Show the answer and scoring guide
$\displaystyle \frac{120}{360} \cdot \pi (15)^2 = \frac{1}{3} \cdot 225\pi = 75\pi \approx 235.6$ square feet.
Check off the points you earned:
(b) Find the length of the curved outer edge of the watered region, to the nearest tenth.
Show the answer and scoring guide
$\displaystyle \frac{120}{360} \cdot 2\pi(15) = 10\pi \approx 31.4$ feet.
Check off the points you earned:
(c) If the sprinkler is set to spin all the way around, write the equation of the circle at the edge of its spray. Is a flower bed at $(13, 9)$ inside, on, or outside this circle?
Show the answer and scoring guide
$(x - 4)^2 + (y + 3)^2 = 225$. Distance to $(13, 9)$: $\displaystyle \sqrt{9^2 + 12^2} = \sqrt{225} = 15$, so the flower bed is exactly on the circle.
Check off the points you earned: