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SAT Math · Question of the Day

SAT Math Question of the Day

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Friday, October 23, 2026

A circle has equation x2+y2−4x+10y+13=0x^2+y^2-4x+10y+13=0. What is the center of the circle?

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Question of the Day

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A circle has equation x2+y2−4x+10y+13=0x^2+y^2-4x+10y+13=0. What is the center of the circle?

  1. (−2,−5)(-2,-5)
  2. (2,5)(2,5)
  3. (2,−5)(2,-5) (correct answer)
  4. (−2,5)(-2,5)

Explanation: To find the center of the circle from x2+y2−4x+10y+13=0x^2 + y^2 - 4x + 10y + 13 = 0, we need to complete the square. Rearranging: (x2−4x)+(y2+10y)=−13(x^2 - 4x) + (y^2 + 10y) = -13. For x: (x−2)2−4(x-2)^2 - 4; for y: (y+5)2−25(y+5)^2 - 25. This gives us (x−2)2+(y+5)2−4−25=−13(x-2)^2 + (y+5)^2 - 4 - 25 = -13, which simplifies to (x−2)2+(y+5)2=16(x-2)^2 + (y+5)^2 = 16. The center is (2,−5)(2, -5). A common error is mixing up the signs when identifying the center from the completed square form.