Math Problem Statement

Write the general form of a circle which passes through the points A(2,3)B(6,1)C(4-3)

Solution

To find the general form of the equation of a circle that passes through three points A(2,3)A(2, 3), B(6,1)B(6, 1), and C(4,3)C(4, -3), we follow these steps:

Step 1: General Form of the Circle

The general equation of a circle is: x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0 where DD, EE, and FF are constants.

Step 2: Substitute the Points into the General Form

Substitute the coordinates of the points AA, BB, and CC into the equation to get three equations.

For point A(2,3)A(2, 3): (2)2+(3)2+2D+3E+F=0(2)^2 + (3)^2 + 2D + 3E + F = 0 4+9+2D+3E+F=0(Equation 1)4 + 9 + 2D + 3E + F = 0 \quad \text{(Equation 1)} 2D+3E+F=132D + 3E + F = -13

For point B(6,1)B(6, 1): (6)2+(1)2+6D+1E+F=0(6)^2 + (1)^2 + 6D + 1E + F = 0 36+1+6D+E+F=0(Equation 2)36 + 1 + 6D + E + F = 0 \quad \text{(Equation 2)} 6D+E+F=376D + E + F = -37

For point C(4,3)C(4, -3): (4)2+(3)2+4D3E+F=0(4)^2 + (-3)^2 + 4D - 3E + F = 0 16+9+4D3E+F=0(Equation 3)16 + 9 + 4D - 3E + F = 0 \quad \text{(Equation 3)} 4D3E+F=254D - 3E + F = -25

Step 3: Solve the System of Equations

We now have the following system of linear equations:

2D + 3E + F &= -13 \quad \text{(1)}\\ 6D + E + F &= -37 \quad \text{(2)}\\ 4D - 3E + F &= -25 \quad \text{(3)} \end{align*}$$ Subtract Equation 1 from Equation 2: $$(6D + E + F) - (2D + 3E + F) = -37 - (-13)$$ $$4D - 2E = -24$$ $$2D - E = -12 \quad \text{(4)}$$ Subtract Equation 3 from Equation 2: $$(6D + E + F) - (4D - 3E + F) = -37 - (-25)$$ $$2D + 4E = -12 \quad \text{(5)}$$ ### Step 4: Solve for $$D$$, $$E$$, and $$F$$ From Equation 4: $$E = 2D + 12$$ Substitute into Equation 5: $$2D + 4(2D + 12) = -12$$ $$2D + 8D + 48 = -12$$ $$10D = -60$$ $$D = -6$$ Now, substitute $$D = -6$$ into $$E = 2D + 12$$: $$E = 2(-6) + 12 = 0$$ Substitute $$D = -6$$ and $$E = 0$$ into Equation 1 to find $$F$$: $$2(-6) + 3(0) + F = -13$$ $$-12 + F = -13$$ $$F = -1$$ ### Step 5: Write the Equation Substituting $$D = -6$$, $$E = 0$$, and $$F = -1$$ into the general form: $$x^2 + y^2 - 6x - 1 = 0$$ This is the general form of the circle passing through the points $$A(2, 3)$$, $$B(6, 1)$$, and $$C(4, -3)$$. Would you like to see more details or have any questions? Here are 5 related questions: 1. How do you convert the general form of a circle into its standard form? 2. How can you find the center and radius of a circle from the general form? 3. What is the significance of the coefficients $$D$$, $$E$$, and $$F$$ in the general form of a circle? 4. How do you find the equation of a circle given its center and radius? 5. How do you determine if three points lie on a circle? **Tip:** When solving a system of equations, it's often helpful to eliminate variables step by step to simplify the problem.

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Math Problem Analysis

Mathematical Concepts

Analytical Geometry
Circle Equations
Linear Algebra

Formulas

General equation of a circle
System of linear equations

Theorems

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Suitable Grade Level

Advanced High School