Question:

Given the ends of the diameter of a circle at (-3,11) and (3,-13), what is the standard form of the equation of the circle? 1 =

Given the ends of the diameter of a circle at (-3,11)
and (3,-13), what is the standard form of the equation
of the circle?
1
=
=
O x2 + (y - 3)2 = 153
o (x - 4)2 + (y + 3)2 = 125
o x2 + (y - 3)2 = 13
+
ES
US

Given the ends of the diameter of a circle at (-3,11) and (3,-13), what is the standard form of the equation of the circle? 1 = = O x2 + (y - 3)2 = 153 o (x - 4)2 + (y + 3)2 = 125 o x2 + (y - 3)2 = 13 + ES US

Answer

b.
Ends of diameter of circle are (-3, 11), (3,-13). Then
circle is given by
(x+3)(n-3)(0-10
) (8+13) = 0.
→ X 9 ryutay-143 30
Juozy +1-1-9-143=0.
afget) - 153
20
x
=
-
"-24
>
x

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