Precalculus

Analytic Geometry

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Circle

Definition of a Circle

In geometry, a circle is a two-dimensional, closed and perfectly round shape. It is defined as the set of all points in a plane that are at a fixed distance (called the radius) from a single point, known as the center of the circle. The distance from the center to any point P(x, y) on the circle's boundary is always the same.

Definition of a circle in geometry
Figure 1.

Standard Equation of a Circle

The equation of a circle with radius R centered at the origin has the form

\[x^2 + y^2 = R^2\]

The equation of a circle with radius R centered at the point \(M\left( {a,b} \right)\) is written as

\[\left({x - a}\right)^2 + \left( {y - b} \right)^2 = R^2\]
A circle with radius R centered at the point A(a,b)
Figure 2.

Equation of a Circle in Three-Point Form

The equation of a circle passing through three points (three-point form) is given by

\[\left| {\begin{array}{*{20}{c}} {{x^2+y^2}} & {x} & {y} & {1}\\ {{x_1^2+y_1^2}} & {x_1} & {y_1} & {1}\\ {{x_2^2+y_2^2}} & {x_2} & {y_2} & {1}\\ {{x_3^2+y_3^2}} & {x_3} & {y_3} & {1} \end{array}} \right| = 0\]

Here \(A\left( {{x_1},{y_1}} \right)\), \(B\left( {{x_2},{y_2}} \right)\), \(C\left( {{x_3},{y_3}} \right)\) are three distinct points lying on the circle.

A circle passing through three points
Figure 3.

Equation of a Circle in Parametric Form

\[\left\{ \begin{aligned} x &= R \cos t \\ y &= R\sin t \end{aligned} \right.,\;\; 0 \le t \le 2\pi\]

where \(x\), \(y\) are the coordinates of any point of the circle, \(R\) is the radius of the circle, \(t\) is a parameter.

General Form of a Circle Equation

\[A{x^2} + A{y^2} + Dx + Ey + F = 0\]

provided that \(A \ne 0\) and \(D^2 + E^2 \gt 4AF\).

The equation involves terms up to the second degree in both the \(x\) and \(y\) coordinates. Thus, a circle is a curve of the second order.

The centre of the circle has coordinates \(\left( {a,b} \right)\) where

\[a = -\frac{D}{2A},\;b = -\frac{E}{2A}\]

The radius of the circle is given by

\[R = \sqrt{\frac{D^2 + E^2 - 4AF}{2\left|A\right|}}\]

Solved Problems

Example 1.

Find the radius and coordinates of the center of the circle \({x^2 + y^2 - 2x + 4y + 1 = 0}.\)

Solution.

We write the given equation in standard form by completing the squares:

\[x^2 + y^2 - 2x + 4y + 1 = 0, \Rightarrow x^2 - 2x + 1 + y^2 + 4y + \underbrace{4 - 4}_{0} = 0,\]
\[\Rightarrow \left({x^2 - 2x + 1}\right) + \left({y^2 + 4y + 4}\right) - 4 = 0, \Rightarrow \left({x - 1}\right)^2 + \left({y + 2}\right)^2 = 2^2.\]

As you can see, the radius is \(R = 2\) and the center of the circle has coordinates \(\left({1,-2}\right).\)

Example 2.

Check that two given circles touch each other: \({\left({x - 1}\right)^2 + \left({y - 2}\right)^2 = 2},\) \({\left({x - 5}\right)^2 + \left({y - 6}\right)^2 = 18}.\)

Solution.

The first circle has radius \(R_1 = \sqrt{2}\) and its center is located at \(M\left({1,2}\right).\) The radius of the second circle is \(R_2 = \sqrt{18}\) and its center is at \(N\left({5,6}\right).\) If two circles touch, then the distance between their centers is equal to the sum of their radii. So we need to check that

\[d_{MN} = R_1 + R_2.\]

Calculate the distance between the centers of the circles:

\[d_{MN} = \sqrt{\left({x_N - x_M}\right)^2 + \left({y_N - y_M}\right)^2} = \sqrt{\left({5 - 1}\right)^2 + \left({6 - 2}\right)^2} = \sqrt{4^2+4^2} = 4\sqrt{2}.\]

Find now the sum of the radii:

\[R_1 + R_2 = \sqrt{2} + \sqrt{18} = \sqrt{2} + 3\sqrt{2} = 4\sqrt{2}.\]

we see that \({d_{MN} = R_1 + R_2 = 4\sqrt{2}.}\) Therefore, these circles are tangent to each other.

Example 3.

Find the equation of a circle passing through the points \(A\left({1,3}\right),\) \(B\left({10,0}\right),\) and \(O\left({0,0}\right).\)

Solution.

–°ompose the determinant using the coordinates of these three points:

\[\left| {\begin{array}{*{20}{c}} {{x^2+y^2}} & {x} & {y} & {1}\\ {{x_1^2+y_1^2}} & {x_1} & {y_1} & {1}\\ {{x_2^2+y_2^2}} & {x_2} & {y_2} & {1}\\ {{x_3^2+y_3^2}} & {x_3} & {y_3} & {1} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {{x^2+y^2}} & {x} & {y} & {1}\\ {{1^2+3^2}} & {1} & {3} & {1}\\ {{10^2+0^2}} & {10} & {0} & {1}\\ {{0^2+0^2}} & {0} & {0} & {1} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {{x^2+y^2}} & {x} & {y} & {1}\\ {{10}} & {1} & {3} & {1}\\ {{100}} & {10} & {0} & {1}\\ {{0}} & {0} & {0} & {1} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {{x^2+y^2}} & {x} & {y}\\ {{10}} & {1} & {3}\\ {{100}} & {10} & {0} \end{array}} \right|.\]

Expand the third-order determinant by the last row:

\[\left| {\begin{array}{*{20}{c}} {{x^2+y^2}} & {x} & {y}\\ {{10}} & {1} & {3}\\ {{100}} & {10} & {0} \end{array}} \right| = 100\left| {\begin{array}{*{20}{c}} {x} & {y}\\ {1} & {3} \end{array}} \right| - 10\left| {\begin{array}{*{20}{c}} {{x^2+y^2}} & {y}\\ {10} & {3} \end{array}} \right| + 0 = 100\left({3x-y}\right) - 10\left({3x^2 + 3y^2 - 10y}\right) = 300x - \cancel{100y} - 30x^2 - 30y^2 + \cancel{100y} = -30\left({x^2 + y^2 - 10x}\right).\]

So the circle equation has the form

\[-30\left({x^2 + y^2 - 10x}\right) = 0, \Rightarrow x^2 + y^2 - 10x = 0, \Rightarrow x^2 - 10x + 25 + y^2 = 25, \Rightarrow \left({x - 5}\right)^2 + y^2 = 5^2.\]

Example 4.

Write the equation of a circle given its diameter end points \(A\left({-5,-1}\right)\) and \(B\left({3,5}\right).\)

Solution.

The diameter of the circle is

\[d_{AB} = \sqrt{\left({x_B - x_A}\right)^2 + \left({y_B - y_A}\right)^2} = \sqrt{\left({3 - \left(-5\right)}\right)^2 + \left({5 - \left(-1\right)}\right)^2} = \sqrt{8^2 + 6^2} = \sqrt{100} = 10.\]

Find the center \(M\left({x_M,y_M}\right)\) of the circle by dividing the segment \(AB\) in half:

\[x_{M} = \frac{x_A + x_B}{2} = \frac{-5+3}{2} = -1;\;\;y_{M} = \frac{y_A + y_B}{2} = \frac{-1+5}{2} = 2.\]
A circle given by the diameter endpoints
Figure 4.

It is clear that the radius is

\[R = AM = MB = \frac{AB}{2} = 5.\]

Hence the circle equation is written as

\[\left({x - x_M}\right)^2 + \left({y - y_M}\right)^2 = R^2, \Rightarrow \left({x + 1}\right)^2 + \left({y - 2}\right)^2 = 25.\]

Example 5.

Find the distance from point \(N\left({-2,1}\right)\) to the circle \({x^2 + y^2 - 4x - 8y + 11 = 0}.\)

Solution.

Let \(d\) be the distance from the point \(N\) to the center of the circle \(M\left({a,b}\right).\) Then the shortest distance \(\ell\) from the point \(N\) to the circle is given by

\[\ell = d - R,\]

where \(R\) is the radius of the circle.

We convert the general equation of the circle to standard form:

\[x^2 + y^2 - 4x - 8y + 11 = 0, \Rightarrow \left({x^2 - 4x + 4}\right) + \left({y^2 - 8y + 16}\right) - 9 = 0, \]
\[\Rightarrow \left({x - 2}\right)^2 + \left({y - 4}\right)^2 = 3^2.\]

So the center of the circle is at point \(M\left({2,4}\right)\) and the radius of the circle is \(R = 3.\)

Distance from point to circle
Figure 5.

Calculate the distance \(d\) between points \(N\) and \(M:\)

\[d = d_{NM} = \sqrt{\left({x_M - x_N}\right)^2 + \left({y_M - y_N}\right)^2} = \sqrt{\left({2 - \left(-2\right)}\right)^2 + \left({4 - 1}\right)^2} = \sqrt{4^2+3^2} = \sqrt{25} = 5.\]

Then the shortest distance from point N to the circle is

\[\ell = d - R = 5 - 3 = 2.\]

Example 6.

Find the equation of a circle of radius \(R\) touching the coordinate axes.

Solution.

The distance from the center of the given circle to the coordinate axes is equal to the radius. That is, the center of the circle in the first quarter has coordinates \(M\left({R,R}\right).\)

A circle of radius R touching the coordinate axes
Figure 6.

The circle can be located in each quarter. So in total we get four equations of the circle:

\[\left({x \pm R}\right)^2 + \left({y \pm R}\right)^2 = R^2.\]

We can represent these equations in the general form:

\[x^2 \pm 2Rx + R^2 + y^2 \pm 2Ry + \cancel{R^2} = \cancel{R^2}, \Rightarrow x^2 + y^2 \pm 2Rx \pm 2Ry + R^2 = 0.\]