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What is the center of the circle x2+y2−144=0x^2+y^2-144=0? \newlineSimplify any fractions. \newline(□,□)(\square, \square)

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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+y2−144=0x^2+y^2-144=0? \newlineSimplify any fractions. \newline(□,□)(\square, \square)
  1. Write Given Equation: We start by writing the given equation of the circle: x2+y2−144=0x^2 + y^2 - 144 = 0. To find the center, we need to get the equation in the standard form of a circle, 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 and rr is the radius.
  2. Move Constant: Move the constant to the other side of the equation by adding 144144 to both sides.\newlinex2+y2−144+144=0+144x^2 + y^2 - 144 + 144 = 0 + 144\newlineThis simplifies to x2+y2=144x^2 + y^2 = 144.
  3. Compare with Standard Form: Now, we compare the equation x2+y2=144x^2 + y^2 = 144 with the standard form of a circle's equation (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 shift the xx and yy from their original positions.
  4. Find Center: Therefore, the center of the circle given by the equation x2+y2−144=0x^2 + y^2 - 144 = 0 is (h,k)=(0,0)(h, k) = (0, 0).

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