Find the Center and Radius x^2+10x+41+y^2-8y=36

Problem

x2+10*x+41+y2−8*y=36

Solution

1. Rearrange the equation by grouping the x terms and y terms together and moving the constant to the right side.

x2+10*x+y2−8*y=36−41

x2+10*x+y2−8*y=−5

2. Complete the square for the x terms by adding (10/2)2=25 to both sides of the equation.

x2+10*x+25+y2−8*y=−5+25

3. Complete the square for the y terms by adding ((−8)/2)2=16 to both sides of the equation.

x2+10*x+25+y2−8*y+16=−5+25+16

4. Factor the perfect square trinomials and simplify the constant on the right side.

(x+5)2+(y−4)2=36

5. Identify the center (h,k) and the radius r by comparing the result to the standard form of a circle (x−h)2+(y−k)2=r2

h=−5

k=4

r2=36⇒r=6

Final Answer

Center: *(−5,4), Radius: *6

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