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Byju's Answer
Standard XII
Mathematics
Equation of Circle with Origin as Center
The equation ...
Question
The equation o circle passing through
(
3
,
4
)
and center at
(
1
,
2
)
is
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Solution
The equation of circle is
(
x
−
h
)
2
+
(
y
−
k
)
2
=
r
2
.
Center is
(
1
,
2
)
The equation of circle is
(
x
−
1
)
2
+
(
y
−
2
)
2
=
r
2
$
It passes through
(
3
,
4
)
⟹
(
3
−
1
)
2
+
(
4
−
2
)
2
=
r
2
2
2
+
2
2
=
r
2
r
2
=
4
+
4
=
8
So the equation of circle is
(
x
−
1
)
2
+
(
y
−
2
)
2
=
8
x
2
−
2
x
+
1
+
y
2
−
4
y
+
4
=
8
x
2
+
y
2
−
2
x
−
4
y
−
3
=
0
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