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Sagot :
Answer:
The center is located at (−1, −2), and the radius is 5.
Step-by-step explanation:
How to determine the center and the radius?
The circle equation is given as:
x^2 + 2x + y^2 + 4y = 20
Complete the square on x and y.
So, we have:
x^2 + 2x + (2/2)^2+ y^2 + 4y + (4/2)^2 = 20 + (2/2)^2 + (4/2)^2
Evaluate the exponents
x^2 + 2x + 1+ y^2 + 4y + 4 = 20 +1 + 4
Expand
x^2 + x + x + 1+ y^2 + 2y + 2y + 4 = 25
Factorize
x(x + 1) + 1(x + 1)+ y(y + 2) + 2(y + 2) = 25
Factor out x + 1 and y + 2
(x + 1)^2 + (y + 2)^2 = 25
Express 25 as 5^2
(x + 1)^2 + (y + 2)^2 = 5^2
The equation of a circle is represented as:
(x - a)^2 + (y - b)^2 = r^2
Where:
Center = (a,b)
Radius = r
So, we have:
Center = (-1,-2)
Radius = 5
Hence, the center is located at (−1, −2), and the radius is 5.
brainliest please.
[tex]\qquad\qquad\huge\underline{{\sf Answer}}[/tex]
Let's solve ~
We have to use complete the square method to find standard equation of circle and identify the center coordinates and radius of the circle.
Here we go ~
[tex]\qquad \sf \dashrightarrow \: {x}^{2} + 2x + {y}^{2} + 4y = 20[/tex]
[tex]\qquad \sf \dashrightarrow \: {x}^{2} + 2x +1 + {y}^{2} + 4y + 4= 20 + 1 + 4[/tex]
[ Adding 1 and 4 on both sides ]
[tex]\qquad \sf \dashrightarrow \: {(x} + 1) {}^{2} + ({y}^{} + 2) {}^{2} = 25[/tex]
Now, compare the given equation with standard equation ~
[tex]\qquad \sf \dashrightarrow \:(x - h) { }^{2} + (y - k {)}^{2} = r {}^{2} [/tex]
- h = x - coordinate of circle
that is :
[tex]\qquad \sf \dashrightarrow \: - h = + 1[/tex]
[tex]\qquad \sf \dashrightarrow \:h = - 1[/tex]
similarly
- k = y - coordinate of the circle
that is :
[tex]\qquad \sf \dashrightarrow \: - k = + 2[/tex]
[tex]\qquad \sf \dashrightarrow \: k = - 2[/tex]
So, coordinates of centre is (-1 , -2)
And r in the equation is Radius ~
[tex]\qquad \sf \dashrightarrow \: {r}^{2} =2 5[/tex]
[tex]\qquad \sf \dashrightarrow \:r = \sqrt{25} [/tex]
[tex]\qquad \sf \dashrightarrow \: \therefore \: radius = 5 \: \: units[/tex]
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