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Find the center and the radius of the circle with the equation x^2 +2x+y^2-2y14+0

User Victoriah
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2 Answers

3 votes

Answer:

Center is (-1,1)

Radius is 4

Explanation:


x^2+2x+y^2-2y-14=0\\(x^2+2x)+(y^2-2y)=14\\(x^2+2x+1)+(y^2-2y+1)=14+1+1\\(x+1)^2+(y-1)^2=16

Therefore, the center of the circle is
(h,k)=(-1,1) and the radius is 4

User Vincet
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5 votes
To determine the center and the radius of a circle from its equation, we need to rewrite the equation in standard form, which is (x - h)^2 + (y - k)^2 = r^2.

The given equation is x^2 + 2x + y^2 - 2y + 14 = 0.

To rewrite it in standard form, we complete the square for both variables x and y.

For x, we add (2/2)^2 = 1 to both sides of the equation:
x^2 + 2x + 1 + y^2 - 2y + 14 = 1
(x + 1)^2 + y^2 - 2y + 15 = 1

For y, we add (-2/2)^2 = 1 to both sides of the equation:
(x + 1)^2 + (y - 1)^2 + 15 = 1 + 1
(x + 1)^2 + (y - 1)^2 + 15 = 2

Now, the equation is in standard form: (x + 1)^2 + (y - 1)^2 = 2 - 15
(x + 1)^2 + (y - 1)^2 = -13

Comparing it to (x - h)^2 + (y - k)^2 = r^2, we can determine that the center of the circle is at (-1, 1), and the radius is sqrt(-13).

Since the radius is the square root of a negative number, it means that the given equation does not represent a circle in the Euclidean coordinate system.
User Sergey Zhigalov
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9.2k points

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