
To find: the center and radius of each circle and graph the circle.

Answer to Problem 23RE
The center of the circle (h, k) is (1, -2) and the radius r is 3.
Explanation of Solution
Given:
x2+y2−2x+4y−4=0
Calculation:
Group the terms containing x and the terms containing y . Bring the constant to theright side of the equation.
(x2−2x)+(y2+4y)=4
Express the terms within the parentheses as perfect squares. Any number added on the left-hand side must be added on the right-hand side also.
(x2−2x+1)+(y2+4y+4)=4+1+4 ↑︷(22)2=1 ↑︷(42)2=4
Factor the expression.
(x−1)2+(y+2)2=32
The standard equation of a circle with center (h, k) and radius r is (x−h)2+(y−k)2=r2 .
Express the equation in standard form
(x−1)2+(y+2)2=32
Compare the equation with the standard form.
(x−h)2+(y− k)2= r2 ↓ ↓ ↓(x−1)2+(y−(−2))2=32
Therefore, the center of the circle (h, k) is (1, -2) and the radius r is 3.
The center of the circle is at the point (1, -2). Since the radius is 3 units, four points on the circle can be located by plotting points that are 3 units to the right, to the left, up, and down from the center. Use these points to graph the circle.
Substitute y=0 in the standard equation to find the x -intercept.
(x−1)2+(0+2)2= 9(x−1)2+ 4 = 9
Subtract 4 from both the sides.
(x−1)2+4−4=9−4 (x−1)2=5
Use the square root method.
x−1=±√5
Add 1 to both the sides.
x−1+1=±√5+1x=1±√5
The x -intercepts are 1+√5 and 1−√5 .
Substitute x=0 in the standard equation to find the y-intercept
(0−1)2+(y+2)2=91+(y+2)2=9
Subtract 1 from both the sides.
1+(y+2)2−1=9−1(y+2)2=8
Use the square root method.
y+2=±√8
Subtract 2 from both the sides.
y+2−2 = ±√8−2y=−2±√8
The y intercepts are −2+√8 and −2−√8 .
Conclusion:
Therefore, the center of the circle (h, k) is (1, -2) and the radius r is 3.
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