Page 6 - For Okta_Float

Example 3 The equation of a circle is 2x 2y 4x 12y 18 0.
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a. Write the standard form of the equation.
b. Find the radius and the coordinates of the center.
c. Graph the equation.
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a. 2x 2y 4x 12y 18 0
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x y 2x 6y 9 0 Divide each side by 2.
Group to form perfect
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(x 2x ?) (y 6y ?) 9
square trinomials.
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(x 2x 1) (y 6y 9) 9 1 9 Complete the square.
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(x 1) (y 3) 19 Factor the trinomials.
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(x 1) (y 3) (19 ) 2 Express 19 as (19 ) to show
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that r 19 .
b. The center of the circle is located at (1, 3), and y
the radius is 19 .
O x
c. Plot the center at (1, 3). The radius of 19 is (1, 3)
approximately equal to 4.4.
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(x 1) (y 3) 19

From geometry, you know that any two points in the coordinate plane
determine a unique line. It is also true that any three noncollinear points in the
coordinate plane determine a unique circle. The equation of this circle can be
found by substituting the coordinates of the three points into the general form of
the equation of a circle and solving the resulting system of three equations.

Example 4 Write the standard form of the equation of the circle that passes through the
points at (5, 3), (2, 2), and (1, 5). Then identify the center and radius of
the circle.
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Substitute each ordered pair for (x, y) in x y Dx Ey F 0, to create a
system of equations.
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(5) (3) D(5) E(3) F 0 (x, y) (5, 3)
Look Back 2 2
Refer to Lesson 2-2 (2) (2) D(2) E(2) F 0 (x, y) (2, 2)
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to review solving (1) (5) D(1) E(5) F 0 (x, y) (1, 5)
systems of three
equations. Simplify the system of equations. 5D 3E F 34 0
2D 2E F 8 0
D 5E F 26 0
The solution to the system is D 4, E 2, and F 20.
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The general form of the equation of the circle is x y 4x 2y 20 0.
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After completing the square, the standard form is (x 2) (y 1) 25.
The center of the circle is at (2, 1), and its radius is 5.
626 Chapter 10 Conics