How to Find the Equation of a Circle Passing Through Given Points: A Comprehensive Guide

When dealing with circles in geometry, finding the equation of a circle that passes through a set of points can be a useful and practical skill in various fields such as engineering, physics, and mathematics. In this guide, we will explore how to find the equation of a circle passing through the points (0,0), (5,2), and (2,1) using the general equation of a circle and solving a system of equations.

Understanding the General Equation of a Circle

The general equation of a circle is given by:

x2 y2 Dx Ey F 0

To find the equation of a circle passing through specific points, we need to determine the coefficients D, E, and F. Let's go through the process step-by-step with the given points (0,0), (5,2), and (2,1).

Using the Given Points to Formulate Equations

We start by substituting each point into the general equation of the circle to form a system of equations. For each point, we will derive an equation by setting the left-hand side of the circle equation to zero (since the points lie on the circle).

Substitution and Simplification Steps

Point (0,0)

Substituting (0,0) into the circle equation:

02 02 D ? 0 E ? 0 F 0

This simplifies to:

F 0

Point (5,2)

Substituting (5,2) into the circle equation:

52 22 D ? 5 E ? 2 F 0

This simplifies to:

25 4 5D 2E F 0

Using F 0 (from the previous step), we get:

25 4 5D 2E 0

Simplifying further:

5D 2E 29 0 … Equation 1

Point (2,1)

Substituting (2,1) into the circle equation:

22 12 D ? 2 E ? 1 F 0

This simplifies to:

4 1 2D E F 0

Using F 0 (again), we get:

2D E 5 0 … Equation 2

Solving the System of Equations

Now we have a system of two equations with two unknowns (D and E):

5D 2E 29 0 (Equation 1) 2D E 5 0 (Equation 2)

Solving for D and E

We can solve for E in terms of D from Equation 2:

E -2D - 5

Substitute this expression for E into Equation 1:

5D 2(-2D - 5) 29 0

Simplifying:

5D - 4D - 10 29 0

D - 10 29 0

D -19

Now substitute D -19 back into the expression for E:

E -2(-19) - 5 38 - 5 33

Final Equation and Standard Form

The coefficients we have found are D -19 and E 33, and we already know F 0. Thus, the equation of the circle is:

x2 y2 - 19x 33y 0

To rewrite this in standard form, we complete the square for both x and y terms:

x2 - 19x y2 33y 0

Completing the square for x:

x2 - 19x (x - 19/2)2 - (19/2)2 (x - 19/2)2 - 361/4

Completing the square for y:

y2 33y (y 33/2)2 - (33/2)2 (y 33/2)2 - 1089/4

Substituting these back into the equation:

(x - 19/2)2 - 361/4 (y 33/2)2 - 1089/4 0

Combining the constants:

(x - 19/2)2 (y 33/2)2 361/4 1089/4 1450/4 725/2

This represents a circle centered at (19/2, -33/2) with radius √(725/2).

Key Takeaways: The equation of the circle passing through the points (0,0), (5,2), and (2,1) is x2 y2 - 19x 33y 0. After simplification, it can be written as (x - 19/2)2 (y 33/2)2 725/2. The process involves solving a system of equations derived from the general form of the circle equation.

Related Keywords: circle equation system of equations completing the square