Geometry: Circles, Coordinates & Volume

How to do it:

1. Name the arc the angle intercepts (the slice of the circle between the two rays).

2. A central angle has its vertex at the center; it equals the intercepted arc.

3. An inscribed angle has its vertex on the circle; it is half that same arc, so half the central angle.

4. A radius that meets a tangent is perpendicular at the point of contact.

Worked example: A central angle of $80°$ intercepts an arc. An inscribed angle on that same arc is $40°\text{.}$

Check: An inscribed angle on a diameter is a right angle (Thales). If a radius meets a tangent, the angle is $90°\text{,}$ not $45°\text{.}$

How to do it:

1. Every point $(x,y)$ on the circle is $r$ units from the center $(h,k)\text{.}$

2. Write the distance: $\sqrt{(x-h)^2+(y-k)^2}=r\text{.}$

3. Square both sides: $(x-h)^2+(y-k)^2=r^2\text{.}$

4. If you are given $x^2+y^2+Dx+Ey+F=0\text{,}$ complete the square to recover $(h,k)$ and $r\text{.}$

Worked example: Center $(2,-1)\text{,}$ radius 5: $(x-2)^2+(y+1)^2=25\text{.}$

Check: Radius 5 means $r^2=25\text{,}$ not $5\text{.}$ Signs flip: center $(-3,4)$ is $(x+3)^2+(y-4)^2=r^2\text{.}$ See circle (standard form).

How to do it:

1. Read each slope $m$ from $y=mx+b\text{,}$ or compute rise over run.

2. Parallel lines have the same $m$ and a different intercept (unless they are the same line).

3. Perpendicular lines satisfy $m_1 \cdot m_2 = -1\text{,}$ so $m_{\perp}=-1/m\text{,}$ unless a line is vertical.

4. A vertical line has undefined slope and is perpendicular to every horizontal line.

Worked example: The line through $(0,1)$ parallel to $y=2x-4$ is $y=2x+1\text{.}$ A perpendicular line through $(0,1)$ is $y=-\frac{1}{2}x+1\text{.}$

Check: $2 \times (-1/2) = -1\text{.}$ Two lines with slope $2$ are parallel, not perpendicular.

How to do it:

1. Choose a solid (or a stack of solids) that matches the object.

2. Cylinder $V=\pi r^2 h\text{.}$ Cone $V=\tfrac{1}{3}\pi r^2 h\text{.}$ Sphere $V=\tfrac{4}{3}\pi r^3\text{.}$ Pyramid $V=\tfrac{1}{3}Bh\text{.}$

3. Add the pieces. A hemisphere is half a sphere: $\tfrac{2}{3}\pi r^3\text{.}$

4. Keep units and $\pi$ until the problem asks for a decimal.

Worked example: A silo is a cylinder of radius 4 m and height 10 m plus a hemisphere cap. Cylinder $160\pi\text{.}$ Hemisphere $\tfrac{2}{3}\pi(64)=\tfrac{128\pi}{3}\text{.}$ Total $\tfrac{480\pi}{3}+\tfrac{128\pi}{3}=\tfrac{608\pi}{3}$ cubic meters.

Check: Grade 8 volume of solids computed the formulas; here you choose the model. A tree trunk is closer to a cylinder than a sphere (modeling with geometry).

How to do it:

1. Group the $x$ terms and the $y$ terms, and move the constant to the right side.

2. For each group, add $\left(\tfrac{b}{2}\right)^2$ to both sides to make a perfect square.

3. Read off $(h,k)$ and $r$ from $(x-h)^2+(y-k)^2=r^2\text{.}$

Worked example: $x^2+y^2-6x+4y-12=0\text{.}$ Group: $(x^2-6x)+(y^2+4y)=12\text{.}$ Add $9$ and $4$ to both sides: $(x-3)^2+(y+2)^2=25\text{.}$ Center $(3,-2)\text{,}$ radius $5\text{.}$

Check: the point $(3,3)$ is $5$ units above the center, and $9+9-18+12-12=0$ confirms it is on the original circle. Forgetting to add the same numbers to the right side is the usual slip.

How to do it:

1. A radius perpendicular to a chord bisects it, so half the chord, the distance to the center, and the radius form a right triangle.

2. Two tangent segments drawn to a circle from the same outside point are congruent.

3. In a quadrilateral inscribed in a circle, opposite angles are supplementary, because together they intercept the whole circle.

Worked example: a chord of length $16$ sits in a circle of radius $10\text{.}$ Half the chord is $8\text{,}$ so the distance from the center to the chord is $\sqrt{10^2-8^2}=6\text{.}$ In inscribed quadrilateral $ABCD$ with $\angle A=80°\text{,}$ $\angle C=100°\text{.}$

Check: $6$-$8$-$10$ is a scaled $3$-$4$-$5\text{,}$ so the right triangle checks. Two tangent segments from a point $13$ units from the center of a radius-$5$ circle each have length $12\text{,}$ another $5$-$12$-$13\text{.}$

How to do it:

1. Place the figure on a grid, using the origin and the axes to keep the numbers simple.

2. Use slopes to prove parallel or perpendicular, the distance formula to prove congruent lengths, and midpoints to prove a segment is bisected.

3. Say which definition or theorem the numbers satisfy.

Worked example: $A(0,0)\text{,}$ $B(4,0)\text{,}$ $C(4,3)\text{.}$ Slope $AB=0$ and $BC$ is vertical, so the angle at $B$ is right and $\triangle ABC$ is a right triangle. Check with lengths: $4\text{,}$ $3\text{,}$ and $AC=5\text{,}$ a $3$-$4$-$5\text{.}$ For $P(0,0)\text{,}$ $Q(6,0)\text{,}$ $R(8,3)\text{,}$ $S(2,3)\text{,}$ both diagonals have midpoint $(4,1.5)\text{,}$ so $PQRS$ is a parallelogram.

Check: a coordinate proof about all rectangles uses variables such as $(0,0)\text{,}$ $(a,0)\text{,}$ $(a,b)\text{,}$ $(0,b)\text{;}$ specific numbers only prove the one figure. Replay the distance formula.

How to do it:

1. To find the point $P$ that is a fraction $t$ of the way from $A$ to $B\text{:}$ $P=(x_A+t(x_B-x_A),\,y_A+t(y_B-y_A))\text{.}$ A ratio of $1{:}3$ means $t=\tfrac{1}{4}\text{.}$

2. Perimeter: add the distances between consecutive vertices.

3. Area of a right triangle or rectangle comes straight from side lengths; for other polygons, split into triangles and rectangles.

Worked example: $A(1,2)\text{,}$ $B(9,6)\text{,}$ ratio $1{:}3$ from $A\text{:}$ $P=(1+\tfrac{1}{4}\cdot 8,\,2+\tfrac{1}{4}\cdot 4)=(3,3)\text{.}$ The triangle $(0,0)\text{,}$ $(4,0)\text{,}$ $(4,3)$ has perimeter $4+3+5=12$ and area $\tfrac{1}{2}\cdot 4\cdot 3=6\text{.}$

Check: $P$ should lie between $A$ and $B$ and closer to $A\text{;}$ $(3,3)$ does. The midpoint is the special case $t=\tfrac{1}{2}\text{.}$

How to do it:

1. Slice a solid parallel to its base and the cross-section matches the base; slice a cylinder through its axis and you get a rectangle.

2. Cavalieri's principle: two solids with the same height and the same cross-sectional area at every level have the same volume. That is why a slanted stack of coins holds as much as a straight one, and why a cone is one third of its cylinder.

3. Density $=\dfrac{\text{mass}}{\text{volume}}$ (or people per unit area). Multiply a volume by density to get mass.

Worked example: a cylindrical tank of radius $1$ m and height $2$ m holds $\pi(1)^2(2)=2\pi\approx 6.28$ m$^3$ of water; at $1000$ kg per m$^3$ that is about $6{,}280$ kg. A town of $50{,}000$ people on $20$ km$^2$ has a population density of $2{,}500$ people per km$^2\text{.}$

Check: units must match: m$^3$ times kg per m$^3$ gives kg. Slicing a sphere anywhere gives a circle, but the circles shrink toward the poles, so a sphere is not a cylinder.