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Difficulty: HardGeometric Figures on the Coordinate Plane

A circle in the standard (x,y)(x, y) coordinate plane passes through the points (0,0)(0, 0), (8,0)(8, 0), and (2,6)(2, 6). What are the coordinates of the center of this circle?

  1. A
    (3,3)(3, 3)
  2. (4,2)(4, 2)Answer
  3. C
    (4,3)(4, 3)
  4. D
    (5,3)(5, 3)
  5. E
    (2,4)(2, 4)

Answer

The center of the circle is (4,2)(4, 2)
The correct answer is (4,2)(4, 2). A circle's center (h,k)(h, k) is equidistant from all points on the circle. By setting up the distance formula from the center (h,k)(h, k) to the points (0,0)(0,0), (8,0)(8,0), and (2,6)(2,6), we obtain the equations h2+k2=(h−8)2+k2h^2 + k^2 = (h-8)^2 + k^2 and h2+k2=(h−2)2+(k−6)2h^2 + k^2 = (h-2)^2 + (k-6)^2. Solving the first equation yields 16h=6416h = 64, which simplifies to h=4h = 4. Substituting h=4h = 4 into the second equation yields 16+k2=4+(k−6)216 + k^2 = 4 + (k-6)^2. Expanding and simplifying gives 16=40−12k16 = 40 - 12k, which simplifies to 12k=2412k = 24, or k=2k = 2. Therefore, the center of the circle is (4,2)(4, 2).

Step-by-Step Solution

1
Set up the distance formula equations for the center of the circle, (h,k)(h, k), which must be equidistant from the three points on the circle: (0,0)(0, 0), (8,0)(8, 0), and (2,6)(2, 6).
The distances from (h,k)(h, k) to the points are h2+k2\sqrt{h^2 + k^2}, (h−8)2+k2\sqrt{(h-8)^2 + k^2}, and (h−2)2+(k−6)2\sqrt{(h-2)^2 + (k-6)^2}. Since they are radii, we equate them: h2+k2=(h−8)2+k2=(h−2)2+(k−6)2h^2 + k^2 = (h-8)^2 + k^2 = (h-2)^2 + (k-6)^2.
The distance from the center of a circle to any point on its circumference is equal to its radius.
2
Equate the distance to (0,0)(0,0) and the distance to (8,0)(8,0), then solve for hh.
h2+k2=(h−8)2+k2  ⟹  h2=h2−16h+64  ⟹  16h=64  ⟹  h=4h^2 + k^2 = (h-8)^2 + k^2 \implies h^2 = h^2 - 16h + 64 \implies 16h = 64 \implies h = 4.
Since both points lie on the x-axis, the x-coordinate of the center must lie on the perpendicular bisector of the segment connecting them, which is x=4x = 4.
3
Substitute h=4h = 4 into the equation equating the distance to (0,0)(0,0) and (2,6)(2,6) to solve for kk.
42+k2=(4−2)2+(k−6)2  ⟹  16+k2=4+k2−12k+36  ⟹  16=40−12k  ⟹  12k=24  ⟹  k=24^2 + k^2 = (4-2)^2 + (k-6)^2 \implies 16 + k^2 = 4 + k^2 - 12k + 36 \implies 16 = 40 - 12k \implies 12k = 24 \implies k = 2.
Substituting the known coordinate allows us to solve the remaining single-variable equation for the yy-coordinate of the center.
4
Combine the coordinates to state the final center of the circle.
The center is (4,2)(4, 2).
Both coordinates (h,k)(h, k) have been uniquely determined.

Key Concept

Circumcenter of a triangle / Center of a circle passing through three points on a coordinate plane
Estimated Time:2m 0s
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Topics

Geometric Figures on the Coordinate Plane

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