Coordinated Circles

Today I have a very common question - how to find an equation of a circle from three points, which are a requirement for a unique circle. See how this can be done the hard way and the easy way.

Find the center, radius, and area of a circle that passes through (6, 2), (0, -6), and (7, 1).

The usual way of doing this type of problems is finding the equations of two lines between those points, finding their perpendicular bisectors, find their point of intersection (the center), and go from there. There is one very good adjective to describe this: boring. Since this blog is all about interesting math, you'll see there is a way that cuts all this line of operations up to "finding their perpendicular bisectors". Yes, removing two very annoying steps.

Since the center of the circle is equidistant from all three points, we need to find a point which is equidistant from the first two points, then find a point which is equidistant from the second and third point, and then see if they match. If they do, the point we found is equidistant from the first and third point (by the transitive property).

So, let's equate the distance between (6, 2) and (0, -6) (the generic point is (x, y)):
√((x - 6)² + (y - 2)²) = √((x - 0)² + (y + 6)²)

Square both sides:
(x - 6)² + (y - 2)² = x² + (y + 6)²

Open parentheses and add like terms:
x² - 12x + 36 + y² - 4y + 4 = x² + y² + 12y + 36
-12x - 4y + 4 = 12y
-12x - 16y + 4 = 0
-16y = 12x - 4
y = -3x/4 + 1/4

We didn't get a point, we got a line. This line is the perpendicular bisector of the segment between the two points, because the geometric location of all the points equidistant from two points is a line. This line is the perpendicular bisector of the segment connecting the two points. Now you know how to find a perpendicular bisector of two points in one very easy step.

Let's do the same thing to the second pair of points:
√((x - 0)² + (y + 6)²) = √((x - 7)² + (y - 1)²)
x² + (y + 6)² = (x - 7)² + (y - 1)²
x² + y² + 12y + 36 = x² - 14x + 49 + y² - 2y + 1
12y + 36 = -14x + 49 - 2y + 1
14y = -14x + 14
y = -x + 1

Now we have two sets of points that are equidistant from a given set of points. All we need now is to equate those lines and find the point that is equidistant from all three points:
y = -3x/4 + 1/4
y = -x + 1

Equate the y's:
-3x/4 + 1/4 = -x + 1
x/4 = 3/4
x = 3

So:
y = -3 + 1
y = -2

The point which is equidistant from all three points and thus the center of the circle is (3, -2).

Now everything becomes really easy. To find the radius of the circle, we need to find the distance from the center to one of the points. Let's say to (7, 1):
r = √((7 - 3)² + (1 - (-2))²)
r = √(4² + 3²)
r = √25
r = 5

Using the formula for the area of a circle, we finally answer the question completely with the answer:
S = 5² * π = 25π

Problem solved.
Nadav

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Circular Coordinate Geometry

Today I have a question involving some coordinate geometry, or analytic geometry. It also involves some circles. It's quite easy, but somewhat long. Here it comes.

What is the sum of the radii of all circles going through the points (1, 9) and (8, 8) and also tangent to the x-axis?

First, you should know that if a circle is tangent to the x-axis, its radius is the y-coordinate of its center. Let's call the center (x, y), so its radius is y.

Since it's a circle, the distance between the center and each of the points must be y. Let's use the distance formula to show that:

y = sqrt((x - 1)² + (y - 9)²)
y = sqrt((x - 8)² + (y - 8)²)

Square and open parentheses:
y² = x² - 2x + 1 + y² - 18y + 81
y² = x² - 16x + 64 + y² - 16y + 64

y^2 cancels on all equations (I also moved the y terms):
18y = x² - 2x + 82 /*8
16y = x² - 16x + 128 /*9

144y = 8x² - 16x + 656
144y = 9x² - 144x + 1152

9x² - 144x + 1152 = 8x² - 16x + 656
x² - 128x + 496 = 0
(x - 4)(x - 124) = 0
x = 4, 124

Now let's see what's y (the radius) for each x:
16y = 4² - 16*4 + 128 = 80
y = 5

When x = 124:
16y = 124² - 16*124 + 128 = 13520
y = 845

So the two circles have radii of 5 and 845, so the sum of the radii is 850.

Hope you liked it.
Nadav

nadavs