# Math > Math Concepts  > Algebra > Conic Sections >Circle

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## Circle

a

In the following figure we are passing a plane (red cross hatch) from the front to the back of the cone. And we are doing this parallel to the base of the cone (normal to its axis.) The intersection of this plane with the cone is shown in bright red. Now, to see the shape in the plane, imagine rotating the plane toward you over the yellow-green bar. You see a circle with radius r at center C, C is on the axis of the cone.

Now, let's derive an equation of this circle. In all conic sections the X-Y axis in the intersecting plane.

Every point (xp, yp) along the circumference of this circle generates a rectangle (shaded) with corners on the X axis, the Y axis, the point itself , and the center C. The radius r splits this rectangle into two congruent right triangles. Using the Pythagorean Theorem xp, yp and r are related by the equation

xp2 + yp2 = r2

Since this is true for every point on the circle, we can drop the subscript p and the general equation for a circle centered at the origin is therefore

 x2 + y2 = r2

Consider the points on the circle where the circle intersects the X and Y axes (the blue and the red points shown below).

 X axis:The red points are the intersections with the X axis. Here the Y values are 0. The points are: (-r, 0) and (r , 0) Y axis:The blue points are the intersections with the Y axis. Here the X values are 0. The points are: (0, r) and (0, -r) These points represent the extreme values for x and y that satisfy the equation for this circle. Notice that one coordinate is zero when the other is ±r.(The general case follows shortly.)

Limiting the x and y values for all points on the circle to ±r isn't enough to ensure the point lies on the circle. The points in the red hatched areas but not on the boundaries of the square (excluding the 4 intersections discussed previously) do not lie on the circle nor do the points within circle lie on the circle, even though magnitudes of both x and y are less than or equal to r. The equation of the circle eliminates these points. Since we are discussing these regions, the equation for all points inside the circle (not on it) would be

x2 + y2 < r2

And the equation for the points in the shaded region would be

(note all 4 conditions must be met)

x2 + y2 > r2   and

|x| <= r   and    |y| <=r   and   (x,y) {(0, ± r ), ( ± r, 0) }

The equation for all points outside the box is:

|x| > r and |y| > r

We have two regions, with areas nearly the same. Maybe we could find a circle and a square whose areas are the same, in other words flatten the circle to create a square whose area is the same as the circle's area. This is the quadrature, or 'squaring the circle' problem, considered by the early Greeks, and proven to be impossible to do. Yet, intuitively, it seems reasonable to do.

Now let's consider a circle whose center is not at the origin.

This graph shows a circle whose center C is at (h,k), and whose radius is r.

Also shown is an arbitrary point (xp, yp) that lies on the circle.

This right triangle has the two legs whose lengths are:

(k – yp) and (xp – h),

Pythagorean's Theorem gives us:

(k – yp)2 + (xp – h)2 = r2

Since this point was chosen arbitrarily, this equation holds for every point along the circle (the radius r is constant) so, we can drop the subscript p and get the

general equation for a circle (noting that (k - yp)2 = (yp - k)2 ):

 (x - h)2 + (y - k)2 = r2

This equation is a translation of the circle from the origin to the point (h,k) as shown in the next figure.

Also note the center of the translated circle is at +h and +k even though the equation shows them as negative.

For the case one or both are negative say C = (-3, 5):

by direct substitution

(x - (-3))2 + (y - 5)2 = r2 h = -3 and k = 5

Typically we would drop the double negative sign and write

(x + 3)2 + (y - 5)2 = r2

You need to master the negative values for h and k in the standard forms of the conic sections.

Now, some examples:

Some observations:

1) The definition of a circle is : All points (x,y) that are the same distance from a third point (xc, yc). Using set notation:

Circle C = { x,y: (x-xc)2 + (y-yc)2 = r2} for constants xc, yc, and positive r

2) For a given area, the circle has a minimal perimeter. Consider a square with area 100. Its perimeter is 40. ( s2 = 100, s = 10, p = 4s = 410 = 40)

For the circle, its perimeter is 35.4.

(100 = π r2 so r = (100/π)½ r = 5.64, now the circumference is

2 π r = 2 π (5.64) = 35.4 )

3) A similar result is the sphere has the minimum surface area of any given volume. One result of this is the sphere has the least heat loss of any other object with the same volume. Geodesic houses are built with this principle using regular n-gons to approximate the sphere.

This is another FREE ALGEBRA PRINTABLE presented to you from the Algebra section of K12math.com