Solve the problem.
-Find the equation of a circle in standard form where $C ( 6 , - 2 )$ and $D ( - 4,4 )$ are endpoints of a diameter.
A) $( x + 1 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 136$
B) $( x + 1 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 34$
C) $( x - 1 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 34$
D) $( x - 1 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 136$

Question

Quizplus · Accepted Answer

The center of the circle is the midpoint of the diameter, which is $(1, 1)$. The radius is half the distance between $C$ and $D$, calculated as $\sqrt{34}$. The equation is $(x - 1)^2 + (y - 1)^2 = 34$.

Solve the Problem C(6,−2)C ( 6 , - 2 )C(6,−2) And D(−4,4)D ( - 4,4 )D(−4,4)

Solve the Problem $C ( 6 , - 2 )$ And $D ( - 4,4 )$