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Analytic Geometry: (lesson 1 of 3)

Circle

circle in coordinate plane

Definition:

An equation of circle of radius rr with a center in a point O(a,b)O (a, b) is:

(xa)2+(yb)2=r2\color{blue}{(x - a)^2 + (y - b)^2 = r^2}.

If the center of a circle coincides with the origin of coordinates, then an equation of the circle is:

x2+y2=r2\color{blue}{x^2 + y^2 = r^2}.

The General Form of the Circle

An equation, written in the following form, represents a circle except when
D2+E2FD^2 + E^2 \le F

x2+y2+2Dx+2Ey+F=0\color{blue}{x^2 + y^2 + 2Dx + 2Ey + F = 0}

This is called the general form of the circle .

C(D,E)C(-D, -E) is the center of the circle, and the radius is r=D2+E2Fr = \sqrt{D^2 + E^2 - F}

Example 1:

Find the radius and center of the circle x2+y22x4y+1=0x^2 + y^2 - 2x - 4y + 1 = 0

Solution:

It is needed to get the equation into the form: (xa)2+(yb)2=r2(x - a)^2 + (y - b)^2 = r^2

x2+y22x4y+1=0x22x+y24y=1x22x+1+y24y+4=1+1+4(x1)2+(y2)2=22 \begin{aligned} &x^2 + y^2 - 2x - 4y + 1 = 0 \\ &x^2 - 2x + y^2 - 4y = -1 \\ &x^2 - 2x + 1 + y^2 - 4y \color{red}{+ 4} = -1 \color{blue}{+ 1} \color{red}{+ 4} \\ &\color{blue}{(x - 1)^2} + \color{red}{(y - 2)^2} = 2^2 \end{aligned}

The radius of circle is r=2r = 2 and the center of the circle is O(1,2)O(1, 2).

In this example D=1D = -1, E=2E = 2, F=1F = 1

Equation of a Circle from 3 Points

Example 2:

Find the equation of the circle through the points A(4,2)A(4, -2), B(6,1)B(6, 1), C(1,3)C(-1, 3).

Let x2+y2+2Dx+2Ex+F=0x^2 + y^2 + 2Dx + 2Ex + F = 0 represent the circle. Since AA is on the circle, its coordinates, 4 and -2, satisfy the equation

42+(2)2+2D4+2E(2)+F=04^2 + (-2)^2 + 2D \cdot 4 + 2E \cdot (-2) + F = 0

Whence: 8D4E+F=208D - 4E + F = -20.

Similarly, for BB, 12D+2E+F=3712D + 2E + F = -37.

and for CC, 2D+6E+F=10-2D + 6E + F = -10.

Solving this, gotten values are: D=2310D = - \frac{23}{10}, E=1310E = - \frac{13}{10}, F=345F = -\frac{34}{5}, and the equation is: x2+y2235x135y345=0x^2 + y^2 - \frac{23}{5} x - \frac{13}{5} y - \frac{34}{5} = 0

Equation of a tangent at a given point

Let A(x1,y1)A(x_1, y_1) be a point of the circle $(x - a)^2 + (y - b)^2 = r^2 , then an equation of tangent line to circle is:

(x1a)(xa)+(y1b)(yb)=r2\color{blue}{(x_1 - a) (x - a) + (y_1 - b) (y - b) = r^2}

Example:

Given the circle (x1)2+(y2)2=25(x - 1)^2 + (y - 2)^2 = 25 and the point A(4,6)A(4,6) on the circle, find the equation of the tangent to the circle at AA.

Solution:

Here we have: a=1a = 1, b=2b = 2, x1=4x_1 = 4, y1=6y_1 = 6

The equation of tangent is:

(x1a)(xa)+(y1b)(yb)=f2(41)(x1)+(62)(y2)=523(x1)+4(y2)=253x+4y3825=03x+4y36=0 \begin{aligned} (x_1 - a) (x - a) + (y_1 - b) (y - b) &= f^2 \\ (4 - 1) (x - 1) + (6 - 2) (y - 2) &= 5^2 \\ 3 (x - 1) + 4 (y - 2) &= 25 \\ 3x + 4y - 3 - 8 - 25 &= 0 \\ \color{blue}{3x + 4y - 36} &\color{blue}{= 0} \end{aligned}