Write the standard form of the equation of the circle which has (-5, 4) and (-1, 8) as endpoints of a diameter.
A) $( x + 3 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = \sqrt { 8 }$
B) $( x - 3 ) ^ { 2 } + ( y + 6 ) ^ { 2 } = 8$
C) $( x - 3 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 8$
D) $( x - 3 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = \sqrt { 8 }$
E) $( x + 3 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 8$

Question

Write the standard form of the equation of the circle which has (-5, 4) and (-1, 8) as endpoints of a diameter.  
A) $( x + 3 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = \sqrt { 8 }$
B) $( x - 3 ) ^ { 2 } + ( y + 6 ) ^ { 2 } = 8$
C) $( x - 3 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 8$
D) $( x - 3 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = \sqrt { 8 }$
E) $( x + 3 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = 8$

Quizplus · Accepted Answer

The center of the circle is the midpoint of the diameter, which is calculated as $\left(\frac{-5 - 1}{2}, \frac{4 + 8}{2}\right) = (-3, 6)$. The radius is the distance between the center and one of the endpoints, which can be calculated using the distance formula. The distance (radius) squared is $((-5 + 3)^2 + (4 - 6)^2 = 8)$. Thus, the equation of the circle in standard form is $(x + 3)^2 + (y - 6)^2 = 8$.

Write the Standard Form of the Equation of the Circle (x+3)2+(y−6)2=8( x + 3 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = \sqrt { 8 }(x+3)2+(y−6)2=8​

Write the Standard Form of the Equation of the Circle $( x + 3 ) ^ { 2 } + ( y - 6 ) ^ { 2 } = \sqrt { 8 }$