Solve the Problem $( x + 1 ) ^ { 2 } + ( y + 1 ) ^ { 2 } = 136$

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) $( x - 1 ) ^ { 2 } + ( y - 1 ) ^ { 2 } = 136$

Correct Answer:

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