Q. A circle in the xy-plane has the equation x2+y2−14y−51=0. What is the center of the circle?Choose 1 answer:(A) (51,14)(B) (7,10)(C) (0,0)(D) (0,7)
Complete the square for y terms: First, we need to complete the square for the y terms. x2+y2−14y=51To complete the square, take half of the coefficient of y, square it, and add it to both sides.(−14/2)2=49x2+y2−14y+49=51+49
Rewrite with completed square: Now, rewrite the equation with the completed square. x2+(y−7)2=100
Identify center of circle: The equation x2+(y−7)2=100 is now in the form of a standard circle equation (x−h)2+(y−k)2=r2, where (h,k) is the center of the circle.So, h=0 and k=7.
Center corresponds to answer: The center of the circle is (0,7), which corresponds to answer choice (D).
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