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

Anne L.
Hsiao-harng S.
Benjamin T.