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Difficulty: MediumEquations and Graphs of Circles

In the standard (x,y)(x, y) coordinate plane, the endpoints of a diameter of a circle are (−2,5)(-2, 5) and (4,−3)(4, -3). Which of the following is the equation of this circle?

  1. (x−1)2+(y−1)2=25(x - 1)^2 + (y - 1)^2 = 25Answer
  2. B
    (x+1)2+(y+1)2=25(x + 1)^2 + (y + 1)^2 = 25
  3. C
    (x−1)2+(y−1)2=100(x - 1)^2 + (y - 1)^2 = 100
  4. D
    (x−1)2+(y−1)2=5(x - 1)^2 + (y - 1)^2 = 5
  5. E
    (x+1)2+(y+1)2=100(x + 1)^2 + (y + 1)^2 = 100

Answer

(x−1)2+(y−1)2=25(x - 1)^2 + (y - 1)^2 = 25
The center (h,k)(h, k) of the circle is the midpoint of the diameter endpoints (−2,5)(-2, 5) and (4,−3)(4, -3), which is (1,1)(1, 1). The radius rr is the distance from the center (1,1)(1, 1) to one endpoint (4,−3)(4, -3), giving r=(4−1)2+(−3−1)2=5r = \sqrt{(4-1)^2 + (-3-1)^2} = 5. Thus, r2=25r^2 = 25. Substituting h=1h=1, k=1k=1, and r2=25r^2=25 into standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 yields (x−1)2+(y−1)2=25(x - 1)^2 + (y - 1)^2 = 25.

Step-by-Step Solution

1
Find the center of the circle using the midpoint formula
Center (h,k)=(−2+42,5+(−3)2)=(1,1)(h, k) = \left(\frac{-2 + 4}{2}, \frac{5 + (-3)}{2}\right) = (1, 1)
The center of a circle is the midpoint of any of its diameters.
2
Calculate the radius of the circle using the distance formula
r=(4−1)2+(−3−1)2=32+(−4)2=25=5r = \sqrt{(4 - 1)^2 + (-3 - 1)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5
The radius is the distance from the center to any point on the circle, including the endpoints of the diameter.
3
Write the standard equation of the circle (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2
(x−1)2+(y−1)2=52  ⟹  (x−1)2+(y−1)2=25(x - 1)^2 + (y - 1)^2 = 5^2 \implies (x - 1)^2 + (y - 1)^2 = 25
Substitute h=1h = 1, k=1k = 1, and r=5r = 5 into the standard form of a circle's equation.

Key Concept

Equation of a circle given diameter endpoints
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Topics

Equations and Graphs of Circles

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