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Find the center of the circle.\newlinex2+y2−4=0x^{2}+y^{2}-4=0

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Full solution

Q. Find the center of the circle.\newlinex2+y2−4=0x^{2}+y^{2}-4=0
  1. Circle Equation Transformation: The equation of the circle is given by x2+y2−4=0x^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=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center of the circle and rr is the radius.
  2. Isolating x2x^2 and y2y^2: First, we need to isolate the x2x^2 and y2y^2 terms on one side of the equation. We can do this by adding 44 to both sides of the equation to get x2+y2=4x^2 + y^2 = 4.
  3. Comparing with Standard Form: Now, we compare the equation x2+y2=4x^2 + y^2 = 4 with the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2. We can see that hh and kk must both be 00 because there are no terms to indicate a shift from the origin for xx or yy. Therefore, the center of the circle is at (h,k)=(0,0)(h, k) = (0, 0).

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