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What is the center of the circle x2+y2−121=0x^2 + y^2 - 121 = 0?\newlineSimplify any fractions.\newline(_,_)(\_,\_)

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Find properties of circles from equations in general formright chevron icon

Full solution

Q. What is the center of the circle x2+y2−121=0x^2 + y^2 - 121 = 0?\newlineSimplify any fractions.\newline(_,_)(\_,\_)
  1. Move Constant Other Side: x2+y2−121=0x^2 + y^2 - 121 = 0 Move the constant to the other side. Add 121121 on both sides. x2+y2−121+121=0+121x^2 + y^2 - 121 + 121 = 0 + 121 x2+y2=121x^2 + y^2 = 121
  2. Add Constant Both Sides: x2+y2=121x^2 + y^2 = 121\newlineFind the values of hh and kk compared to (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2.\newlineStandard form of a circle: (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2\newlineComparing the standard form and x2+y2=121x^2 + y^2 = 121, we see that:\newlineh=0h = 0\newlinek=0k = 0
  3. Find Values Compared: We found:\newlineh=0h = 0\newlinek=0k = 0\newlineWhat is the center of the circle x2+y2−121=0x^2 + y^2 - 121 = 0?\newlineSubstitute 00 for hh and 00 for kk in (h,k)(h,k).\newlineCenter of circle: (0,0)(0,0)

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