A circle has a diameter with endpoints at \( (-2,5) \) and \( (4,1) \). What is the equation of the circle? Select all that apply. A) \( (x-1)^{2}+(y-3)^{2}=13 \) B) \( (x-3)^{2}+(y-3)^{2}=13 \) C) \( (x-1)^{2}+(y-3)^{2}=18 \) D) \( x^{2}+y^{2}-2 x-6 y=8 \) E) \( x^{2}+y^{2}-6 x-6 y=34 \) F) \( x^{2}+y^{2}-2 x-6 y=3 \)
Added by Dennis P.
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The center of a circle that has a diameter with endpoints at \((-2,5)\) and \((4,1)\) can be found by averaging the x-coordinates and the y-coordinates of the endpoints. This gives us the center \((C)\) of the circle. \[ C_x = \frac{-2 + 4}{2} = 1 \] \[ C_y = Show more…
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