Answer:
(i) Center = (-1, -2)
(ii) Radius = 5 units
General Circle Equation:
Where:
Rewriting the equation:
[tex]\sf x^2 + 2x + y^2+ 4y = 20[/tex]
[tex]\sf x^2 + 2x + 1^2 - 1^2 + y^2 + 4y + 2^2 - 2^2 = 20[/tex]
[tex]\sf (x + 1)^2 - 1 + (y + 2)^2 -4 = 20[/tex]
[tex]\sf (x + 1)^2 + (y + 2)^2 = 20 + 5[/tex]
[tex]\sf (x + 1)^2 + (y + 2)^2 = 25[/tex]
[tex]\sf (x -(- 1))^2 + (y -(- 2))^2 = 5^2 \quad \leftarrow \ \bf General \ Circle \ Equation[/tex]
Identify the following:
Answer:
Completing the square: Circles
Add the square of half the coefficients of both first degree terms (x and y) to both sides:
[tex]\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}[/tex]
Factor the two trinomials on the left side of the equation:
[tex]\begin{aligned} \implies x^2+2x+1+y^2+4y+4 & = 25\\\\ \implies (x+1)^2+(y+2)^2 & = 25 \end{aligned}[/tex]
Equation of a circle: [tex](x-a)^2+(y-b)^2=r^2[/tex]
(where (a, b) is the center and r is the radius)
Comparing constants:
[tex]\displaystyle (x-a)^2+(y-b)^2=r^2\\\\\phantom{(((((}\downarrow \phantom{(((((((((} \downarrow \phantom{(((((} \downarrow \\\\(x+1)^2+(y+2)^2=25[/tex]
Therefore:
[tex]-a=1 \implies a=-1[/tex]
[tex]-b=2 \implies b=-2[/tex]
[tex]r^2=25 \implies r=\sqrt{25}=5[/tex]
Conclusion:
