What is the equation of this circle with the end points of the diameter are at (-4,-1) and (0,-4)?

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What is the equation of this circle with the end points of the diameter are at (-4,-1) and (0,-4)?

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Answer 1 · 2016-03-14T14:25:15

${(x + 2)}^{2} + {(y + \frac{5}{2})}^{2} = \frac{25}{4}$ $( x + 2 )^2 + ( y + 5/2 )^2 = 25/4$

Explanation:

The midpoint of the diameter is the center C.
So, C is $(\frac{- 4 + 0}{2}, \frac{- 1 - 4}{2}) = (- 2, - \frac{5}{2})$ $( (-4 + 0 )/2, (-1-4)/2) = ( -2, -5/2)$ .
$R a d i u s = \frac{d i a m e t e r}{2} = . \frac{\sqrt{16 + 9}}{2} = \frac{5}{2}$ $Radius = (diameter)/2 = .sqrt( 16 + 9 )/2 = 5/2$
The equation is ${(x + 2)}^{2} + {(y + \frac{5}{2})}^{2} = \frac{25}{4}$ $( x + 2 )^2 + ( y + 5/2 )^2 = 25/4$

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What is the equation of this circle with the end points of the diameter are at (-4,-1) and (0,-4)?

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