Question

100 POINTS! Using the following equation, find the center and radius of the circle. You must show and explain all work and calculations to receive credit. Be sure to leave your answer in exact form.

Answer 1

Answer:

Center: (-4, 1)

Radius: √2

Explanation:

Circle general equation is represented by (x - h)² + (y - k)² = r²

where (h, k) is the center and "r" resembles radius

To write in general equation form, start completing square:

x² + y² + 8x -2y + 15 = 0

x² + 8x + y² - 2y + 15 = 0

(x + 4)² - 4² + (y - 1)² - (-1)² + 15 = 0

(x + 4)² + (y - 1)² - 16 - 1 + 15 = 0

(x + 4)² + (y - 1)² - 2 = 0

General equation form:

(x - (-4))² + (y - 1)² = (√2)²

From here, comparing with the formula equation.

It is determined: center: (-4, 1) and radius: √2

Answer 2

Answer:

center = (-4, 1)

radius = √2

Step-by-step explanation:

Equation of a circle

$(x-a)^2+(y-b)^2=r^2$

where:

• (a, b) is the center
• r is the radius

Given equation:

$x^2+y^2+8x-2y+15=0$

Subtract 15 from both sides to move the constant to the right side of the equation:

$\implies x^2+y^2+8x-2y=-15$

Rearrange the left side to collect like variables:

$\implies x^2+8x+y^2-2y=-15$

Add the square of half the coefficients of the variables x and y to both sides:

$\implies x^2+8x+\left(\dfrac{8}{2}\right)^2+y^2-2y+\left(\dfrac{-2}{2}\right)^2=-15+\left(\dfrac{8}{2}\right)^2+\left(\dfrac{-2}{2}\right)^2$

$\implies x^2+8x+16+y^2-2y+1=2$

Factor the two perfect square trinomials on the left side:

$\implies (x+4)^2+(y-1)^2=2$

Compare the found equation with the general form to find the center and radius of the circle:

• center = (-4, 1)
• radius = √2

Learn more about circle equations here:

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