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What is the center of the circle x2+y2−81=0x^2 + y^2 - 81 = 0?\newlineSimplify any fractions.\newline(____,____)

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Q. What is the center of the circle x2+y2−81=0x^2 + y^2 - 81 = 0?\newlineSimplify any fractions.\newline(____,____)
  1. Rewrite equation to standard form: We start by rewriting the equation to resemble 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.\newlinex2+y2−81=0x^2 + y^2 - 81 = 0\newlineAdd 8181 to both sides to isolate the x2x^2 and y2y^2 terms.\newlinex2+y2=81x^2 + y^2 = 81
  2. Add 8181 to both sides: Now, we compare the equation x2+y2=81x^2 + y^2 = 81 with the standard form of a circle's equation. We notice that the x2x^2 and y2y^2 terms do not have any coefficients or constants added to them, which means that hh and kk are both 00. Therefore, the center of the circle is at (h,k)=(0,0)(h,k) = (0,0).

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