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# x^2+y^2+2x-8y+1=0

find the center, radius, and intercepts of the circle with the given equation and then sketch the graph

x2 + y2 + 2x - 8y + 1 = 0

We have to complete the squares. Rewrite the equation:

x2 + 2x         + y2 - 8y         + 1 = 0

for the x part we have to add 1 to both sides of the equation and for the y part we have to add 16 to both sides:

x2 + 2x + 1 + y2 - 8y + 16 + 1 = + 1 + 16

x2 + 2x + 1 + y2 - 8y + 16 = 16

(x + 1)2 +(y - 4)2 = 16

This is a circle with center in C(-1,4) and radius r = 4.

The formulas we will use: 1. a2 ± 2b + c = (a ± b)2 2. (x-a)2 + (y-b)2 = r2 , (a,b) coordinate of a center and "r" is a radius of the circle. 3. ax2 + bx + c = 0 , x1,2 = (-b ± √(b2 - 4ac))/2a .
~~~~~~~~~~
x2 + y2 + 2x - 8y + 1=0 ----> (x2 + 2x + 1) + (y2 - 2*4*y + 42) - 42 = 0
↓                         ↓
(x + 1)2     +        (y - 4)2         = 42
The center of a circle is (-1,4) and radius is 4 .
x-intercept (I will use the given equation): x2 + 02 + 2x - 8*0 + 1 = 0
(x + 1)2 = 0
±(x + 1) = 0
x1,2 = -1 that's mean x-axis is a tangent for the circle
y-intercept: 02 + y2 +2*0 -8y + 1 = 0
y2 - 8y + 1 = 0
y1,2 = (8 ± √(64 - 4)) / 2 = (8 ± √60)/2 ---> y1 = 4 + √15
y2 = 4 - √15

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