Unit, Lesson. Deriving the Equation of a Circle The graph of an equation in and is the set of all points (, ) in a coordinate plane that satisf the equation. Some equations have graphs with precise geometric descriptions. For eample, the graph of the equation = + is the line with a slope of, passing through the point (0, ). This geometric description uses the familiar concepts of line, slope, and point. The equation = + is an algebraic description of the line. In this lesson, ou will investigate how to translate between geometric descriptions and algebraic descriptions of circles. You have alread learned how to use the Pthagorean Theorem to find missing dimensions of right triangles. Now, ou will see how the Pthagorean Theorem leads us to the distance formula, which leads us to the standard form of the equation of a circle. The standard form of the equation of a circle is based on the distance formula. The distance formula, in turn, is based on the Pthagorean Theorem. The Pthagorean Theorem states that in an right triangle, the square of the hpotenuse is equal to the sum of the squares of the legs. hpotenuse leg leg leg hp leg If r represents the distance between the origin and an point (, ), then + = r. can be positive, negative, or zero because it is a coordinate. can be positive, negative, or zero because it is a coordinate. r cannot be negative because it is a distance. A circle is the set of all points in a plane that are equidistant from a reference point in that plane, called the center. The set of points forms a -dimensional curve that has a central angle of 0. The center of a circle is the point in the plane of the circle from which all points on the circle are equidistant. The center is in the interior of the circle. The radius of a circle is a line segment that etends from the center to a point on the circle. Its length is half of the diameter. For a circle with center (0, 0) and radius r, a point (, ) is on that circle if and onl if + = r. MUnitLesson. //09
The distance formula is used to find the distance between an two points on a coordinate plane. The distance formula states that the distance d between A (, ) and B (, ) is ) ( ) (. The distance formula is based on the Pthagorean Theorem. Look at this diagram of a right triangle with vertices A, B, and C. The distance d between points A and B is unknown. The worked eample that follows shows how the distance formula is derived from the Pthagorean Theorem, using the points from the diagram to find d: AB d d BC AC Pthagorean Theorem Substitute values for sides AB, BC, and AC of the triangle ( ) ( ) Simplif. All squares are nonnegative ) ( ) d ( Take the square root of each side of the equation to arrive at the distance formula For a circle with center (h, k) and radius r, an point (, ) is on the circle if and onl if ( h) ( k) r. Squaring both sides of this equation ields the standard form of an equation of a circle with center (h, k) and radius r : ( h) + ( k) = r. MUnitLesson. //09
If a circle has center (0, 0), then its equation is ( 0) ( 0) r, or r. If the center and radius of a circle are known, then either of the following two methods can be used to write an equation for the circle: Appl the Pthagorean Theorem to derive the equation. Or, substitute the center coordinates and radius directl into the standard form. The general form of an equation of a circle is A + B + C + D + E = 0, where A = B, A 0, and B 0. If an one of the following three sets of facts about a circle is known, then the other two can be determined: center (h, k) and radius r standard equation: ( h) + ( k) = r general equation: A + B + C + D + E = 0 The general form of the equation of a circle comes from epanding the standard form of the equation of the circle. The standard form of the equation of a circle comes from completing the squares from the general form of the equation of a circle. Ever perfect square trinomial has the form binomial: b b b. b b because it is the square of a Completing the square is the process of determining the value of b form the perfect square trinomial, b. b and adding it to + b to Common Errors/Misconceptions confusing the radius with the square of the radius forgetting to square half the coefficient of when completing the square neglecting to square the denominator when squaring a fraction MUnitLesson. //09
Eample Derive the standard equation of the circle with center (0,0) and radius.. Sketch the circle and label the radius and a point (, ) on the circle. (, ). Use the Pthagorean Theorem to derive the standard equation. In order to use the Pthagorean Theorem, there must be a right triangle. To create a right triangle, draw a line from point (, ) that is perpendicular to the -ais. Label the resulting sides of the triangle and. (, ) Substitute the values for each side of the triangle into the formula for the Pthagorean Theorem, leg leg hp. leg leg hp Pthagorean Theorem Substitute values from the triangle Simplif The standard equation of this circle is. MUnitLesson. //09
Eample Derive the standard equation of the circle with center (, ) and radius. Then use a graphing calculator to graph our equation.. Sketch the circle and label a point on the circle (, ). (, ). Use the Pthagorean Theorem to derive the standard equation. Create a right triangle. Draw a vertical segment down from the point (, ) and a horizontal segment left from point (, ) to create the triangle. Draw the radius. The length of the base of the triangle is equal to the absolute value of the difference of the -coordinates of the endpoints. The height of the triangle is equal to the absolute value of the difference of the -coordinates of the endpoints. (, ) MUnitLesson. //09
Eample (continued) Substitute the resulting values for the sides of the triangle into the Pthagorean Theorem. leg leg hp Pthagorean Theorem Substitute values from the triangle ( ) ( ) All squares are nonnegative, so replace the absolute value smbols with parentheses ( ) ( ) Simplif The standard equation of this circle is ( ) ( ).. Solve the standard equation for to obtain functions that can be graphed. ( ) ( ) Standard equation ( ) ( ) Subtract ( ) from both sides ( ) ( ) Add to both sides to solve for. Now graph the two functions, ( ) and ( ). On a TI-8/9: Step : Press [Y=]. Step : At Y, tpe in [] [+] [ ] [] [-] [ ( ] [X, T,, n ] [-] [] [)] [ ] [ ) ] Step : At Y, tpe in [] [-] [ ] [] [-] [ ( ] [X, T,, n ] [-] [] [)] [ ] [ ) ] Step : Press [ WINDOW ] to change the viewing window Step : At Xmin, enter [ (-) ] [ ] Step : At Xma, enter [ ] Step 7: At Xscl, enter [ ] Step 8: At Ymin, enter [ (-) ] [ ] Step 9: At Yma, enter [ ] Step 0: At Yscl, enter [ ] Step : Press [ GRAPH ] MUnitLesson. //09
Eample Write the standard equation and the general equation of the circle that has center (, ) and passes through (, ).. Sketch the circle.. Use the distance formula to find the radius, r. ) ( ) d ( Distance formula r (( ) ( )) ( ) Substitute (, ) and (, ) for (,) and (, ) r ( ) ( ) Simplif r r r r 0. Substitute the center and radius directl into the standard form. ( h) ( k ) r Standard form ( ( )) ( ) Substitute values into the equation, using center ( ) ( ) () ( ) ( ) 0 The standard equation is ( ) ( ) 0. (, ), and the radius r. Square the binomials and rearrange terms to obtain the general form. ( ) ( ) 0 Standard equation ( )( ) ( )( ) 0 Epand the factors 9 0 Square the binomials to obtain trinomials 0 0 Combine the constant terms on the left side 0 0 Subtract 0 from both sides to get 0 on the right side 0 0 Rearrange terms in descending order to obtain the general equation The general equation is 0 0 MUnitLesson. 7 //09
Eample Find the center and radius of the circle described b the equation + 8 + + = 0.. Rewrite the equation in standard form. 8 0 General form of the equation 8 Subtract from both sides to get the constant term on one side 8 Group same-variable terms Net, complete the square for both variables. Add the same values to both sides of the equation, as shown below: 8 8 8 8 ( ) () ( ) () Simplif the equation 8 ( ) ( ) Write the perfect square trinomials as squares of binomials The standard equation is ( ) ( ). Determine the center and radius. ( ) ( ) Standard equation from Step ( ) ( ( )) Rewrite to match the form ( h) ( k ) r For the equation ( h) + ( k) = r, the center is (h, k) and the radius is r, so for the equation ( ) ( ( )) r., the center is (h,k ) (, ) and the radius is MUnitLesson. 8 //09
Eample Find the center and radius of the circle described b the equation + + 0 0 + = 0.. Rewrite the equation in standard form. 0 0 0 General form of the equation 0 9 0 Divide each term on both sides b to make the leading coefficient 0 9 Subtract 9 from both sides to get the constant term on one side 0 9 Group same-variable terms Net, complete the square for both variables. Add the same values to both sides of the equation: 0 0 0 9 00 00 0 9 Simplif the equation 00 0 9 0 9 Write the perfect square trinomials as squares of binomials 9. The standard equation is. Determine the center and radius. 9 Write the standard equation from Step Rewrite to match the form ( h) + ( k) = r For the equation ( h) + ( k) = r, the center is (h, k) and the radius is r, so for the the center is ( h,k ), and the radius is equation, r. MUnitLesson. 9 //09