# Circle

A circle is a geometric figure commonly used in Euclidean geometry. $[asy]unitsize(2cm);draw(unitcircle,blue);[/asy]$ A basic circle.

## Definition

A circle is defined as the set (or locus) of points in a plane with an equal distance from a fixed point. The fixed point is called the center and the distance from the center to a point on the circle is called the radius.

### Coordinate Definition

Using the traditional definition of a circle, we can find the general form of the equation of a circle on the coordinate plane given its radius, $r$, and center $(h,k)$. We know that each point, $(x,y)$, on the circle which we want to identify is a distance $r$ from $(h,k)$. Using the distance formula, this gives $\sqrt{(x - h)^2 + (y - k)^2} = r$ which is more commonly written as $$(x - h)^2 + (y - k)^2 = r^2.$$

Example: The equation $(x - 3)^2 + (y + 6)^2 = 25$ represents the circle with center $(3,-6)$ and radius 5 units. ## Circumference and Area

Given a circle of radius $r$, the circumference (distance around a circle) is $2 \pi r$ and the area is $\pi r^2$. Both formulas involve the mathematical constant pi ( $\pi$).

### Archimedes' Proof of Area

We shall explore two of the Greek mathematician Archimedes demonstrations of the area of a circle. The first is much more intuitive.

Archimedes envisioned cutting a circle up into many little wedges (think of slices of pizza). Then these wedges were placed side by side as shown below: As these slices are made infinitely thin, the little green arcs in the diagram will become the blue line and the figure will approach the shape of a rectangle with length $r$ and width $\pi r$ thus making its area $\pi r^2$.

Archimedes also came up with a brilliant proof of the area of a circle by using the proof technique of reductio ad absurdum.

Archimedes' actual claim was that a circle with radius $r$ and circumference $C$ had an area equivalent to the area of a right triangle with base $C$ and height $r$. First let the area of the circle be $A$ and the area of the triangle be $T$. We have three cases then.

Case 1: The circle's area is greater than the triangle's area.

Case 2: The triangle's area is greater than the circle's area.

Case 3: The circle's area is equal to the triangle's area.

Assume that $A>T$. Let $P$ be the area of a regular polygon that is closest to the circle's area. Therefore we have $A-P so $P>T$. Let the apothem be $a$ and the perimeter be $p$ so the area of a regular polygon is one half of the product of the perimeter and apothem. The perimeter is less than the circumference so $p<2\pi r$ and the apothem is less than the radius so $a. Therefore $P=\frac{1}{2}ap<\frac{1}{2}r\cdot 2\pi r=T$. However it cannot be both $P>T$ and $P. So $A\not >T$.

### Area Proof Using Calculus

Let the circle in question be $x^2 + y^2 = r^2$, where r is the circle's radius. By symmetry, the circle's area is four times the area in the first quadrant. The area in the first quadrant can be computed using a definite integral from 0 to r of the function $f(x) = \sqrt{r^2 - x^2}$. Using the substitution $x = r \sin u, dx = r \cos u$ gives the indefinite integral as $\frac{r^2}{2} (u - \frac{\sin 2u}{2}) + C$, so the definite integral equals $\frac{r^2}{2} * \frac{\pi}{2}$. Multiplying by four gives the area of the circle as $\pi r^2$.

## Lines in Circles $[asy]draw(unitcircle);draw((-0.8,1)--(1,1),Arrow);draw((1,1)--(-0.8,1),Arrow);draw((0,1)--(1,0));[/asy]$

A line that touches a circle at only one point is called the tangent of that circle. Note that any point on a circle can have only one tangent.

A line segment that has endpoints on the circle is called the chord of the circle. If the chord is extended to a line, that line is called a secant of the circle. The longest chord of the circle is the diameter; it passes through the center of the circle.

When two secants intersect on the circle, they form an inscribed angle.

### Properties

• The measure of an inscribed angle is always half the measure of the central angle with the same endpoints.
• Since the diameter divides the circle into two equal parts, any angle formed by the two endpoints of a diameter and a third distinct point on the circle as the vertex is a right angle.
• Also, a right triangle inscribed in a circle has a hypotenuse that is a diameter of the circle.
• Similarly, if a tangent line and a secant line intersects at the point of tangency, the measure of the angle formed is always half the measure of the central angle with the same endpoints.
• From that property, the angle formed by the diameter and a tangent line with the point of tangency on the diameter is a right angle.
• The perpendicular line through the tangent where it touches the circle is a diameter of the circle.
• The perpendicular bisector of a chord is always a diameter of the circle.
• When two chords $AB$ and $CD$ intersect at point $P$ inside the circle, $\angle APC = \frac{m\widehat{AC} + m\widehat{BD}}{2}$.
• When two chords $AB$ and $CD$ intersect at point $P$ outside the circle, $\angle APC = \frac{m\widehat{AC} - m\widehat{BD}}{2}$.
• Lengths of chords can be calculated by using the Power of a point theorem.

## Problems

### Introductory

• Under what constraints is the circumference (in inches) of a circle greater than its area (in square inches)?

### Intermediate

• Circles with centers $A$ and $B$ have radii 3 and 8, respectively. A common internal tangent intersects the circles at $C$ and $D$, respectively. Lines $AB$ and $CD$ intersect at $E$, and $AE=5$. What is $CD$? $$\mathrm{(A) \ } 13\qquad\mathrm{(B) \ } \frac{44}{3}\qquad\mathrm{(C) \ } \sqrt{221}\qquad\mathrm{(D) \ } \sqrt{255}\qquad\mathrm{(E) \ } \frac{55}{3}\qquad$$

(Source)

• Let $$S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\}$$

and $$S_2=\{(x,y)|\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)\}$$. What is the ratio of the area of $S_2$ to the area of $S_1$? $$\mathrm{(A) \ } 98\qquad \mathrm{(B) \ } 99\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 101\qquad \mathrm{(E) \ } 102$$

(Source)

• Consider a circle $S$, and a point $P$ outside it. The tangent lines from $P$ meet $S$ at $A$ and $B$, respectively. Let $M$ be the midpoint of $AB$. The perpendicular bisector of $AM$ meets $S$ in a point $C$ lying inside the triangle $ABP$. $AC$ intersects $PM$ at $G$, and $PM$ meets $S$ in a point $D$ lying outside the triangle $ABP$. If $BD$ is parallel to $AC$, show that $G$ is the centroid of the triangle $ABP$.