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

Question

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

 

Answer 1

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. 

 

Answer 2

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 , x_1,2 = (-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
                                                           x_1,2 = -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
   y_1,2 = (8 ± √(64 - 4)) / 2 = (8 ± √60)/2 ---> y₁ = 4 + √15
                                                                    y₂ = 4 - √15