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A circle in the 
xy-plane has the equation 
x^(2)+y^(2)-14 y-51=0. What is the center of the circle?
Choose 1 answer:
(A) 
(51,14)
(B) 
(7,10)
(C) 
(0,0)
(D) 
(0,7)

A circle in the xy x y -plane has the equation x2+y214y51=0 x^{2}+y^{2}-14 y-51=0 . What is the center of the circle?\newlineChoose 11 answer:\newline(A) (51,14) (51,14) \newline(B) (7,10) (7,10) \newline(C) (0,0) (0,0) \newline(D) (0,7) (0,7)

Full solution

Q. A circle in the xy x y -plane has the equation x2+y214y51=0 x^{2}+y^{2}-14 y-51=0 . What is the center of the circle?\newlineChoose 11 answer:\newline(A) (51,14) (51,14) \newline(B) (7,10) (7,10) \newline(C) (0,0) (0,0) \newline(D) (0,7) (0,7)
  1. Complete the square for y terms: First, we need to complete the square for the y terms. \newlinex2+y214y=51x^2 + y^2 - 14y = 51\newlineTo complete the square, take half of the coefficient of y, square it, and add it to both sides.\newline(14/2)2=49(-14/2)^2 = 49\newlinex2+y214y+49=51+49x^2 + y^2 - 14y + 49 = 51 + 49
  2. Rewrite with completed square: Now, rewrite the equation with the completed square. x2+(y7)2=100x^2 + (y - 7)^2 = 100
  3. Identify center of circle: The equation x2+(y7)2=100x^2 + (y - 7)^2 = 100 is now in the form of a standard circle equation (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center of the circle.\newlineSo, h=0h = 0 and k=7k = 7.
  4. Center corresponds to answer: The center of the circle is (0,7)(0, 7), which corresponds to answer choice (D)(D).

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