MA.912.GR.7.2

Benchmark Instructional Guide

Connecting Benchmarks/Horizontal Alignment

• MA.912.GR.3.3 
• MA.912.GR.5.3
• MA.912.GR.5.4
• MA.912.GR.5.5 
• MA.912.GR.6

Terms from the K-12 Glossary

• Circle 
• Diameter 
• Radius

Vertical Alignment

Previous Benchmarks

• MA.8.GR.1.1
• MA.8.GR.1.2 
• MA.912.AR.1.2 
• MA.912.AR.3.7 
• MA.912.F.2

Next Benchmarks

• MA.912.GR.7

Purpose and Instructional Strategies

• Instruction includes students deriving the equation of a circle centered at the origin and at a point ($h$, $k$). Students should make the connection to the Pythagorean Theorem by drawing right triangles from the center of the circle. Students should make the connection between the equations centered at the origin and at a point ($h$, $k$); realizing that if the point ($h$, $k$) is the origin then the equation can be written as either $x$² + $y$² = $r$² or as ($x$ − 0)² + ($y$ − 0)² = $r$². 
  • For example, given a circle centered at the origin on the coordinate plane and have students select a point $P$($x$, $y$) on the circle. Students should be able to see that the distance from $P$ to the center is the radius of the circle. Students can then draw a right triangle, like the one shown below, that includes the center and $P$ as two of its vertices. Students can use their knowledge of the Pythagorean Theorem to develop the equation $x$² + $y$² = $r$², showing the relationship between $x$, $y$ and the radius, $r$. Students should be able to realize that this equation is valid for every point P($x$, $y$) on the circle and is not valid for any point not on the circle. 
    
    
    
  • For example, given a circle centered at $C$($h$, $k$) on the coordinate plane and have students select a point $P$($x$, $y$) on the circle. Students should be able to see that the distance from $P$ to the center is the radius of the circle. Students can then draw a right triangle, like the one shown below, that includes the center and $P$ as two of its vertices. Students can use their knowledge of the Pythagorean Theorem to develop the equation ($x$ − $h$)² + ($y$ − $k$)² = $r$², showing the relationship between $x$, $y$, $h$, $k$ and the radius, $r$. Students should be able to realize that this equation is valid for every point $P$($x$, $y$) on the circle and is not valid for any point not on the circle. 
    
    
    
• Instruction includes the connection to transformation of functions (as was done in Algebra 1) when developing an understanding of the relationship between the equations $x$² + $y$² = $r$² and ($x$ − $h$)² + ($y$ − $k$)² = $r$². 
• Students should have experience working with the equation of a circle in center-radius form, ($x$ − $h$)² + ($y$ − $k$)² = $r$², and in general form, $A$$x$² + $B$$x$² + $C$$x$ + $D$$y$ + $E$ = 0 where $A$ and $B$ are the same real, non-zero number. Students may need to convert from one form to another, particularly from the general form to the center-radius form. Instruction includes discussing how this conversion can be done and the benefits of each form of the equation of a circle. 
• Instruction includes exploring the pattern of completing the square for the general form and the way the center and the radius can be obtained from this procedure. Exploration includes cases where the radius may result in a complex number; noting that the equation does not give a circle when this happens. 
  • For example, given the equation $A$$x$² + $B$$y$² + $C$$x$ + $D$$y$ + $E$ = 0 when $A$ = 1 and $B$ = 1, students can rewrite this as $x$² + $C$$x$ + $y$² + $D$$y$ = −$E$. Then students can complete the square for both the $x$-part of the equation and the $y$-part of the equation. Students should obtain the equation ($x$ + $\frac{\text{<math><mi>C</mi></math>}}{\text{2}}$)² + ($y$ + $\frac{\text{<math><mi>D</mi></math>}}{\text{2}}$)² = −$E$ + ($\frac{\text{<math><mi>C</mi></math>}}{\text{2}}$)² + ($\frac{\text{<math><mi>E</mi></math>}}{\text{2}}$)². Students should then compare this to center-radius form to determine that the center of the circle is at the point (−$\frac{\text{<math><mi>C</mi></math>}}{\text{2}}$, −$\frac{\text{<math><mi>D</mi></math>}}{\text{2}}$) and the radius is . 
• Problem types include cases where $A$ and $B$ are equivalent, but not equal to 1. 
• For enrichment of this benchmark, instruction includes the connection to finding the equation of the circumscribed circle about a triangle whose vertices are given on the coordinate plane. (MTR.5.1)

Common Misconceptions or Errors

• Students may confuse the signs of ($h$, $k$) when determining the center of the circle.
  • For example, given the equation ($x$ − 1)² + ($y$ + 2)² = 1, the center is (1, −2) and not (−1, 2). • Students may forget to take the square root of the $r$² term to determine the radius of the circle.

Instructional Tasks

• A circle on the coordinate plane is given. Segments $C$$B$ and $E$$D$ are diameters of circle $A$. Point $C$ is located at (3,5), point $D$ is located at (6,6), point $B$ is located at (7,3) and point $E$ is located at (4,2). 
  
  
  
  • Part A. Determine the center of the circle. Explain your method. 
  • Part B. Find the length of the radius of circle $A$. Explain your method. 
  • Part C. Write the equation of the circle and check that the points $B$, $C$, $D$ and $E$ satisfy the equation. 

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