Question

PLEASE HELP ASAP!! Complete the square to rewrite the following equation. Identify the center and radius of the circle. You must show all work and calculations to receive credit.
x2 + 2x + y2 + 4y = 20

Answer 1

Answer:

(i) Center = (-1, -2)

(ii) Radius = 5 units

General Circle Equation:

• (x - h)² + (y - k)² = r²

Where:

• (h, k) is the center points

• r denotes the radius

Rewriting the equation:

$\sf x^2 + 2x + y^2+ 4y = 20$

$\sf x^2 + 2x + 1^2 - 1^2 + y^2 + 4y + 2^2 - 2^2 = 20$

$\sf (x + 1)^2 - 1 + (y + 2)^2 -4 = 20$

$\sf (x + 1)^2 + (y + 2)^2 = 20 + 5$

$\sf (x + 1)^2 + (y + 2)^2 = 25$

$\sf (x -(- 1))^2 + (y -(- 2))^2 = 5^2 \quad \leftarrow \ \bf General \ Circle \ Equation$

Identify the following:

• (h, k) = (-1, -2), radius = 5 units

Answer 2

Answer:

Completing the square:  Circles

Add the square of half the coefficients of both first degree terms (x and y) to both sides:

$\begin{aligned}\implies x^2+2x+\left(\dfrac{2}{2}\right)^2+y^2+4y+\left(\dfrac{4}{2}\right)^2 & =20+\left(\dfrac{2}{2}\right)^2+\left(\dfrac{4}{2}\right)^2\\\\x^2+2x+1+y^2+4y+4 & = 20+1+4\\\\x^2+2x+1+y^2+4y+4 & = 25\end{aligned}$

Factor the two trinomials on the left side of the equation:

$\begin{aligned} \implies x^2+2x+1+y^2+4y+4 & = 25\\\\ \implies (x+1)^2+(y+2)^2 & = 25 \end{aligned}$

Equation of a circle:  $(x-a)^2+(y-b)^2=r^2$

(where (a, b) is the center and r is the radius)

Comparing constants:

$\displaystyle (x-a)^2+(y-b)^2=r^2\\\\\phantom{(((((}\downarrow \phantom{(((((((((} \downarrow \phantom{(((((} \downarrow \\\\(x+1)^2+(y+2)^2=25$

Therefore:

$-a=1 \implies a=-1$

$-b=2 \implies b=-2$

$r^2=25 \implies r=\sqrt{25}=5$

Conclusion:

• center = (-1, -2)
• radius = 5 units