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Our next entry in the conic sections menagerie is the circle. Recall from Geometry that a circle can be determined by fixing a
point (called the center) and a positive number (called the radius) as follows.
A circle with center \((h,k)\) and radius \(r>0\) is the set of all points \((x, y)\) in the plane whose distance to \((h,k)\) is \(r\).
From the diagram, we see that a point \((x,y)\) is on the circle if and only if its distance to \((h,k)\) is \(r\). We express this relationship
algebraically using the Distance Formula, Equation distanceformula, as
\[r = \sqrt {(x - h)^2 + (y-k)^2}\]
By squaring both sides of this equation, we get an equivalent equation (since \(r > 0\)) which gives us the standard equation of a
circle.
The Standard Equation of a Circle:
The equation of a circle with center \((h,k)\) and radius \(r >0\) is \((x-h)^2 + (y-k)^2 = r^2.\)
Note in the standard equation of a circle, both of the variables squared. This is a quick way to distinguish the equation of a
circle from that of a parabola in which only one of the variables is squared.
We put Equation standardcircle to good use in the following example.
\(~\)
For each of the equations below:
Graph the equation in the \(xy\)-plane.
Find the center and radius.
\((x+2)^2+(y-1)^2 = 4\)
\(3x^2 - 6x + 3y^2 + 4y -4 = 0\)
Graph \(f(x) = - \sqrt {4x-x^2}\).
Find the standard form of the circle satisfying the following characteristics:
The points \((-1,3)\) and \((2,4)\) are the endpoints of a diameter.
The circle whose graph is below.
Rewriting \((x+2)^2+(y-1)^2 = 4\) as \((x-(-2))^2+(y-1)^2 = (2)^2\), we identify \(h = -2\), \(k=1\) and \(r = 2\). Thus we have a circle centered at \((-2,1)\) with a radius of \(2\).
To help us create a detailed graph, we start from the center \((-2,1)\) and move two units to the left and two units
up and down to the right to identify four points on the graph. (Note the center of the circle is not on
the graph of the circle!) We get \((-2-2, 1) = (-4,1)\), \((-2+2, 1) = (0,1)\), \((-2, 1+2) = (-2,3)\) and \((-2,1-2) = (-2,-1)\). Our graph is below.
In order to make use of Equation standardcircle, we need to put \(3x^2 - 6x + 3y^2 + 4y -4 = 0\) into standard form. To that end, we complete the square on
both the \(x\) and \(y\) terms and collect the constants to the other side of the equation as demonstrated
below.
From Equation standardcircle, we identify \(h = 1\), \(k = - \frac {2}{3}\), and \(r = \frac {5}{3}\). Hence, we have a circle with center \(\left (1, -\frac {2}{3}\right )\) and radius \( \frac {5}{3}\).
As above, we find four points on the circle by starting at the center \(\left (1, -\frac {2}{3}\right )\) and moving up, down, to the left, and to the
right \( \frac {5}{3}\) units. Doing so produces the following points: \(\left (-\frac {2}{3}, -\frac {2}{3}\right )\) , \(\left (\frac {8}{3}, -\frac {2}{3}\right )\) , \(\left (1, 1\right )\), and \(\left (1, -\frac {7}{3}\right )\). Our graph is below.
We are asked to graph \(f(x) = - \sqrt {4x-x^2}\), which, at first glance, seems out of place in this section. However, to graph means we graph
the equation \(y = -\sqrt {4x-x^2}\).
Squaring both sides, we get \(y^2 = (-\sqrt {4x-x^2})^2\) or \(y^2 = 4x - x^2\). Rearranging this equation gives \(x^2-4x+y^2 = 0\). Completing the square, we obtain \((x-2)^2 + y^2 = 4\) which, when
rewritten as \((x-2)^2 + (y-0)^2 = (2)^2\) is precisely the standard form of a circle as written in Equation standardcircle.
With \(h = 2\), \(k=0\) and \(r=2\), we know the graph of \((x-2)^2 + y^2 = 4\) is a circle of radius \(2\) centered at \((2,0)\). However, the graph we want isn’t the entire circle.
(For one thing, the graph of a circle fails the Vertical Line Test so it does not represent \(y\) as a function of \(x\) in
this case.) Indeed, we want the graph of \(y = -\sqrt {4x-x^2}\). Because of the ‘\(-\) ’, we want the lower semicircle, graphed
below.
We recall that a diameter of a circle is a line segment containing the center and two points on the circle.
The desmos interactive plots the data below and walks us through the steps required to find our solution
geometrically.
The task before us is to analytically follow these steps to find our answer.
First, note that since the given points are endpoints of a diameter, we know their midpoint \((h, k)\) is the center of the
circle. Using Equations midpointformula, we find the center of the circle below.
Likewise, the diameter of the circle is the distance between the given points, so we can find the radius of the circle
by taking half of this distance. Using distanceformula, we get:
From the graph given to us, we are safe to assume the center of the circle is \((-2,2)\) since the circle appears to be
tangent to the coordinate axes at \((-2,0)\) and \((0,2)\). (Recall that for every point \(P\) on the circle, the tangent line at
\(P\) is perpendicular to the radial line containing the center and \(P\). Since the circle is tangent to the
\(x\)-axis at \((-2,0)\), the center must lie on a line perpendicular to the \(x\)-axis which contains \((-2,0)\) or \(x = -2\). Likewise, the
circle is tangent to the \(y\)-axis at \((0,2)\), the center must lie on \(y=2\). Hence the center is \((-2,2)\).) Moreover, since
the distance from \((-2,2)\) to either of \((-2,0)\) or \((0,2)\) is \(2\), the radius of the circle is \(2\). Per Equation standardcircle, our answer is \((x-(-2))^2 + (y-2)^2 = (2)^2\) or
\((x+2)^2+(y-2)^2 = 4\).
In number ctscircleex above, we needed to transform a given equation into the standard form as stated in Equation standardcircle. We record these
steps below. Note that given an equation that represents a circle, both variables need to be squared and the squared terms
must have the same coefficients.
To Write the Equation of a Circle in Standard Form
Group common variables together on one side of the equation and put the constant on the other.
Complete the square on both variables as needed.
Divide both sides by the coefficient of the squares. (For circles, they will be the same.)
It is possible to obtain equations like \((x-3)^2 + (y+1)^2 = 0\) or \((x-3)^2 + (y+1)^2 = -1\), neither of which describes a circle. (Do you see why not?) The reader is encouraged
to think about what, if any, points lie on the graphs of these two equations.
We close this section with a brief discussion of the so-called Unit Circle. (Widely regarded as the most important circle in
all of mathematics.)
The Unit Circle is the circle centered at \((0,0)\) with a radius of \(1\). The standard equation of the Unit Circle is \(x^2 + y^2 = 1.\)
In some ways, we may think of the Unit Circle as the progenitor of all circles. Indeed, if we divide both sides of Equation standardcircle by \(r^2\),
we obtain the alternate standard form of a circle below.
The Alternate Standard Equation of a Circle: The equation of a circle with center \((h,k)\) and radius \(r >0\) is
Our last example has us find some important points on the the Unit Circle.
Find the points on the unit circle with \(y\)-coordinate \(\frac {\sqrt {3}}{2}\).
Note that all points \((x,y)\) on the Unit Circle satisfy the equation \(x^2+y^2 = 1\). Hence, our first step is to replace \(y\) with \(\frac {\sqrt {3}}{2}\) and solve for
\(x\).
PreCalculus![[Picture]](Circles.online-100d4847947a6137f9063941035a2765.svg)
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