Recall that an equation of a circle can be written in standard form (x-h)² + (y-k)² = r, where (h, k) is the center and r is the radius. After expanding both squares, moving all terms to the left-hand side, and combining like terms, the equation can also be written in the form x² + y² + Ax+By+C =0, where A, B, and C are constants. a. Find an equation of the form x² + y² + Ax+By+C =0 for the circle that passes through the points (6, 0), (2, 2), and (7,-3). To do so, find the values of A, B, and C by writing and solving a system of 3 linear equations. System: Show the steps of solving the system. Equation in Standard Form: Solution: b. Rewrite the equation found in part (a) in standard form using the technique of completing the square. c. Use your result in part (b) to determine the center and radius of the circle. Contor A = B = C=

College Algebra
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ISBN:9781337282291
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Chapter4: Rational Functions And Conics
Section4.4: Translations Of Conics
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### Understanding the Equation of a Circle

In this educational module, we will explore how to determine the equation of a circle in standard form and how to convert the general form of the circle equation into the standard form.

#### Recalling the Standard Form of a Circle Equation
The equation of a circle can be written in standard form as:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius.

#### Converting to General Form
After expanding both squares, moving all terms to the left-hand side, and combining like terms, the equation can also be written in the form:
\[ x^2 + y^2 + Ax + By + C = 0 \]
where \(A\), \(B\), and \(C\) are constants.

### Problem-Solving Example

**Part (a): Writing and Solving a System of Equations**

**Problem Statement:**
Find an equation of the form \( x^2 + y^2 + Ax + By + C = 0 \) for the circle passing through the points \((6, 0)\), \((2, 2)\), and \((7, -3)\).

To do this, determine the values of \(A\), \(B\), and \(C\) by writing and solving a system of 3 linear equations.

**Steps for Solving the System:**

1. Substitute the coordinates of each point into the general form equation.
   
   For point \((6, 0)\):
   \[ 6^2 + 0^2 + 6A + 0B + C = 0 \]
   \[ 36 + 6A + C = 0 \]  → (Equation 1)

   For point \((2, 2)\):
   \[ 2^2 + 2^2 + 2A + 2B + C = 0 \]
   \[ 4 + 4 + 2A + 2B + C = 0 \]
   \[ 8 + 2A + 2B + C = 0 \]  → (Equation 2)

   For point \((7, -3)\):
   \[ 7^2 + (-3)^2 + 7A - 3B
Transcribed Image Text:### Understanding the Equation of a Circle In this educational module, we will explore how to determine the equation of a circle in standard form and how to convert the general form of the circle equation into the standard form. #### Recalling the Standard Form of a Circle Equation The equation of a circle can be written in standard form as: \[ (x - h)^2 + (y - k)^2 = r^2 \] where \((h, k)\) is the center and \(r\) is the radius. #### Converting to General Form After expanding both squares, moving all terms to the left-hand side, and combining like terms, the equation can also be written in the form: \[ x^2 + y^2 + Ax + By + C = 0 \] where \(A\), \(B\), and \(C\) are constants. ### Problem-Solving Example **Part (a): Writing and Solving a System of Equations** **Problem Statement:** Find an equation of the form \( x^2 + y^2 + Ax + By + C = 0 \) for the circle passing through the points \((6, 0)\), \((2, 2)\), and \((7, -3)\). To do this, determine the values of \(A\), \(B\), and \(C\) by writing and solving a system of 3 linear equations. **Steps for Solving the System:** 1. Substitute the coordinates of each point into the general form equation. For point \((6, 0)\): \[ 6^2 + 0^2 + 6A + 0B + C = 0 \] \[ 36 + 6A + C = 0 \] → (Equation 1) For point \((2, 2)\): \[ 2^2 + 2^2 + 2A + 2B + C = 0 \] \[ 4 + 4 + 2A + 2B + C = 0 \] \[ 8 + 2A + 2B + C = 0 \] → (Equation 2) For point \((7, -3)\): \[ 7^2 + (-3)^2 + 7A - 3B
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