Deriving the Equation of a Circle Passing Through Given Points and Determining Its Properties

In this article, we will derive the equation of a circle that passes through three given points and explore the geometric properties of the resulting circle. Specifically, we will work with the points A(1, 7), B(-2, -2), and C(-8, 10).

Step 1: Setting Up the General Form of the Circle's Equation

The general equation of a circle is given by:

x2 y2 Dx Ey F 0,

where D, E, and F are constants that need to be determined. We will substitute each of the given points into this equation to find these constants.

Step 2: Substituting the Points into the Equation

For point A(1, 7):

12 72 D(1) E(7) F 0

1 49 D 7E F 0

D 7E F -50 (Equation 1)

For point B(-2, -2):

(-2)2 (-2)2 D(-2) E(-2) F 0

4 4 - 2D - 2E F 0

-2D - 2E F -8 (Equation 2)

For point C(-8, 10):

(-8)2 102 D(-8) E(10) F 0

64 100 - 8D 10E F 0

-8D 10E F -164 (Equation 3)

Step 3: Solving the System of Equations

We now have the following system of equations to solve:

D 7E F -50
-2D - 2E F -8
-8D 10E F -164

To eliminate F from the equations, we subtract Equation 1 from Equation 2 and Equation 1 from Equation 3:

(-2D - 2E F) - (D 7E F) -8 50
-3D - 9E 42
D 3E -14 (Equation 4)

(-8D 10E F) - (D 7E F) -164 50
-9D 3E -114
3D - E 12 (Equation 5)

Step 4: Solving Equations 4 and 5 for D and E

From Equation 4:

D 3E -14 (Equation 6)

From Equation 5, we can express D in terms of E using Equation 6:

D -14 - 3E

Substituting this expression into Equation 5:

3(-14 - 3E) - E 12
-42 - 9E - E 12
-10E 54
E -5.4

Substituting E back into Equation 6:

D -14 - 3(-5.4) -14 16.2 2.2

Using Equation 1 to find F:

2.2 7(-5.4) F -50
F -50 - 37.8 -14.4

Thus, we have:

D 2.2, E -5.4, F -14.4

The equation of the circle is:

x2 y2 2.2x - 5.4y - 14.4 0

Geometric Properties of the Circle

To explore the geometric properties of the circle, we will determine if it forms a right triangle and identify the center and radius.

1. **Slope Calculation for Chords AB and AC:**

Slope of chord AB (y2 - y1) / (x2 - x1) (-2 - 7) / (-2 - 1) 3
**Slope of chord AC (y2 - y1) / (x2 - x1) (10 - 7) / (-8 - 1) -1/3**

2. **Checking if ABC is a Right Triangle:**

The product of the slopes of the two chords gives us -1, indicating that the angle between AB and AC is a right angle.

Therefore, ABC is a right triangle with BC as the hypotenuse.

3. **Finding the Center and Radius:**

Since ABC forms a right triangle, the mid-point of the hypotenuse, denoted by O, is the center of the required circle.

The coordinates of the mid-point of BC are:

(-5, -4)

The radius of the circle, AO, is given by:

AO sqrt[(-5 - 1)2 (-4 - 7)2] sqrt[36 121] sqrt[157]

The equation of the circle in standard form is:

(x 5)2 (y 4)2 45

This matches with the derived circle equation from the earlier steps.