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Mathematics
Mathematics
Problem
If one end of a diameter of the circle
2
x
2
+
2
y
2
−
4
x
−
8
y
+
2
=
0
2x^2 + 2y^2 - 4x - 8y + 2 = 0
2
x
2
+
2
y
2
−
4
x
−
8
y
+
2
=
0
is
(
−
1
,
2
)
(-1, 2)
(
−
1
,
2
)
, then the other end of the diameter is a.
(
2
,
1
)
(2, 1)
(
2
,
1
)
b.
(
3
,
2
)
(3, 2)
(
3
,
2
)
c.
(
4
,
3
)
(4, 3)
(
4
,
3
)
d.
(
5
,
4
)
(5, 4)
(
5
,
4
)
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