Properties of Circles
A Circle is a 2-D figure that is formed by a set of points that are at a fixed and equal distance from a fixed point in the plane, the center of the circle, commonly referred to as the Origin. If you can identify one point on the circumference of a circle, \((x, y)\), you are able to determine the points using symmetry.
Characteristics of Circles
| Characteristic | Description | Formula |
|---|---|---|
| Area | The total amount of space covered by the circle; is dependent on the size of the radius | \(xr^2\) |
| Circumference | The distance around the boundary of a circle | \(2πr\) |
| Radius | A line that starts at the origin and touches a point on the circumference | \(r\) or \(\cfrac{d}{2}\) |
| Diameter | The line that passes through the origin and touches 2 points on the circumference directly opposite each other | \(d\) or \(2r\) |
| Arc | A portion of the circumference; generally referred to as a curve | \(\theta r\) |
| Sector | A slice of the circle enclosed by 2 radii and the included arc of a circle | \(\cfrac{\theta r^2}{2}\) |
| Chord | A straight line that joins 2 different points on the circumference | \(2r\sin(\cfrac{\theta}{2})\) |
| Tangent | A line outside the circle that touches the circumference at a specific point | N/A |
| Secant | A line that intersects the circle at 2 distinct points | N/A |
| Segment | A portion of the circle enclosed by a chord and the included arc of a circle | \(\cfrac{r^2(\theta - \sin\theta)}{2}\) |
NOTE: \(\theta\) is always measured in radians when calculating!!
Equation of A Circle
The Equation of A Circle is used to outline both the radius of a circle in addition to the position of its origin. It can be expressed using the following formulas:
\(x² + y² = r²\)
OR
\((x - h)² + (y - k)² = r²\)
- \(x\) represents the x-coordinate
- \(y\) represents the y-coordinate
- \(r\) represents the radius of the circle
- \(h\) represents horizontal shift
- \(k\) represents vertical shift
One of the main uses for this equation is to determine the relation other coordinates have with the circle, whether they lie inside the circle, on the circle, or outside the circle.
- If \(x² + y² = r²\), then the point lies on the circle
- If If \(x² + y² < r²\), then the point lies within the circle
- If If \(x² + y² > r²\), then the point lies outside the circle
Example
Determine if point \((2,5)\) lies on a circle with equation \(x² + y² = 29\)
We can plug the coordinates into the equation and simplify:
\(= (2)² + (5)²\)
\(= 4 + 25\)
\(29 = 29\)
As \(29 = 29\) we can determine that the point \((2,5)\) lies ON the circle.
\((13, 5)\)
We can plug the coordinates into the equation and simplify:
\(= (13 - 6)² + (5 + 4)²\)
\(= (7)² + (9)²\)
\(= 49 + 81\)
\(130 = 130\)
Since \(130 = 130\) we can determine that the point \((13, 5)\) lies ON the circle.
\((3, 2)\)
We can plug the coordinates into the equation and simplify:
\(= (3 - 6)² + (2 + 4)²\)
\(= (-3)² + (6)²\)
\(= 9 + 36\)
\(45 < 130\)
As \(45 < 130\), we can determine that the point \((3, 2)\) lies WITHIN the circle.
\((-4, 6)\)
We can plug the coordinates into the equation and simplify:
\(= (-4 - 6)² + (6 + 4)²\)
\(= (-10)² + (10)²\)
\(= 100 + 100\)
\(200 > 130\)
As \(200 > 130\), we can determine that the point \((-4, 6)\) lies OUTSIDE the circle.
For the chord \(\text{DE}\) where \(\text{D}(-4,3)\) and \(\text{E}(3,4)\):
- Find the coordinates of the midpoint of \(\text{DE}\) where \(\text{D}(-4,3)\) and \(\text{E}(3,4)\)
- Find the slope of chord \(\text{DE}\)
- Verify that \(\text{OM}\) is perpendicular to \(\text{DE}\)
i. We can determine the midpoint of \(\text{DE}\) by using the midpoint formula:
\(\text{midpoint}_{\text{DE}} = \left(\cfrac{3 + (-4)}{2}, \cfrac{4 + 3}{2}\right)\)
\(\text{midpoint}_{\text{DE}} = \left(\cfrac{-1}{2}, \cfrac{7}{2}\right)\)
\(\text{midpoint}_{\text{DE}} = (-0.5, 3.5)\)
Therefore, we can determine that the midpoint of chord \(\text{DE}\) is \(\boldsymbol{(-0.5, 3.5)}\).
ii. We can use the slope formula to find the slope of chord \(\text{DE}\):
\(m_{\text{DE}} = \cfrac{4 - 3}{3 - (-4)}\)
\(m_{\text{DE}} = \cfrac{1}{7}\)
Therefore, we can determine that the slope of chord \(\text{DE}\) is \(\boldsymbol{\cfrac{1}{7}}\).
iii. We can determine that \(\text{OM}\) is perpendicular to \(\text{DE}\) by finding its slope:
\(m_{\text{OM}} = \cfrac{-3.5}{0.5}\)
\(m_{\text{OM}} = -7\)
We can determine that \(\text{OM}\) is perpendicular to \(\text{DE}\) since \(-7\) and \(\cfrac{1}{7}\) are negative reciprocols of each other.
In order to find the centre of the circle, we need to find the Point of Intersection of 2 perpendicular bisectors of the chords.
We can start by determining the midpoints of 2 of the lines:
First, we can determine the midpoint of line \(\text{DE}\):
\(\text{midpoint}_{\text{DE}} = \left(\cfrac{-7}{2}, \cfrac{13}{2}\right)\)
\(\text{midpoint}_{\text{DE}} = (-3.5, 6.5)\)
Next, we can find the midpoint of line \(\text{EF}\):
\(\text{midpoint}_{\text{EF}} = \left(\cfrac{0}{2}, \cfrac{12}{2}\right)\)
\(\text{midpoint}_{\text{EF}} = (0, 6)\)
We can also find the slopes of each of these lines, and the perpendicular slopes by extension:
First, we can find the slope of line \(\text{DE}\):
\(m_{\text{DE}} = \cfrac{-1}{-3}\)
\(m_{\text{DE}} = \cfrac{1}{3}\)
\(⊥ m_{\text{DE}} = -3\)
Next, we can find the slope of line \(\text{EF}\):
\(m_{\text{EF}} = \cfrac{2}{-4}\)
\(m_{\text{EF}} = \cfrac{-1}{2}\)
\(⊥m_{\text{EF}} = 2\)
We can now plug all the values we have determined thus far into their respective linear equations to determine the \(y\)-intercepts, \(b\):
We can first determine the \(y\)-intercept and equation for line \(\text{DE}\):
\(\text{DE}: 6.5 = -3(-3.5) + b\)
\(\text{DE}: 6.5 = 10.5 + b\)
\(\text{DE}: b = 6.5 - 10.5\)
\(\text{DE}: b = -4\)
\(\text{DE}: y = -3x - 4\)
Next, we can determine the \(y\)-intercept and equation for line \(\text{EF}\):
\(\text{EF}: 6 = 2(0) + b\)
\(\text{EF}: b = 6\)
\(\text{EF}: y = 2x + 6\)
To find the Point of Intersection, we first need to subtract the equations from one another in order to find the \(x\)-coordinate:
\(-3x - 4 = 2x + 6\)
\(-3x - 2x = 6 + 4\)
\(\cfrac{-5x}{-5} = \cfrac{10}{-5}\)
\(x = -2\)
We can then plug the \(x\)-coordinate into one of the equations to determine the \(y\)-coordinate:
\(y = 2(-2) + 6\)
\(y = -4 + 6\)
\(y = 2\)
Therefore, we can determine that the Point of Intersection is \(\boldsymbol{(-2, 2)}\).