Find the center of the circle. x^(2)+y^(2)-4=0

Circle Equation Transformation: The equation of the circle is given by x^2 + y^2 - 4 = 0. To find the center, we need to express the equation in the standard form of a circle's equation, which is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius.Isolating x^2 and y^2: First, we need to isolate the x^2 and y^2 terms on one side of the equation. We can do this by adding 4 to both sides of the equation to get x^2 + y^2 = 4.Comparing with Standard Form: Now, we compare the equation x^2 + y^2 = 4 with the standard form (x - h)^2 + (y - k)^2 = r^2. We can see that h and k must both be 0 because there are no terms to indicate a shift from the origin for x or y. Therefore, the center of the circle is at (h, k) = (0, 0).

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Find properties of circles from equations in general form

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Q. Find the center of the circle.

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x^{2}+y^{2}-4=0

Circle Equation Transformation: The equation of the circle is given by $x^2 + y^2 - 4 = 0$ . To find the center, we need to express the equation in the standard form of a circle's equation, which is $(x - h)^2 + (y - k)^2 = r^2$ , where $(h, k)$ is the center of the circle and $r$ is the radius.
Isolating $x^2$ and $y^2$ : First, we need to isolate the $x^2$ and $y^2$ terms on one side of the equation. We can do this by adding $4$ to both sides of the equation to get $x^2 + y^2 = 4$ .
Comparing with Standard Form: Now, we compare the equation $x^2 + y^2 = 4$ with the standard form $(x - h)^2 + (y - k)^2 = r^2$ . We can see that $h$ and $k$ must both be $0$ because there are no terms to indicate a shift from the origin for $x$ or $y$ . Therefore, the center of the circle is at $(h, k) = (0, 0)$ .