Find the equation of the circle passing through
the points (5,7), (6, 6) and (2, -2).
Answers
Answer:
Let the required equation of the circle be
x
2
+
y
2
+
2
g
x
+
2
f
y
+
c
=
0.
.
.
.
(
i
)
Since it passes through each of the points (5, 7), (6, 6) and (2, -2), each one of these points must satisfy (i).
∴
25
+
49
+
10
g
+
14
f
+
c
=
0
⇒
10
g
+
14
f
+
c
+
74
=
0
.
.
.
(
i
i
)
36
+
36
+
12
g
+
12
f
+
c
=
0
⇒
12
g
+
12
f
+
c
+
72
=
0
.
.
.
(
i
i
i
)
4
+
4
+
4
g
−
4
f
+
c
=
0
⇒
4
g
−
4
f
+
c
+
8
=
0
.
.
.
(
i
v
)
Subtracting (ii) from (iii), we get
2
g
−
2
f
−
2
=
0
⇒
g
−
f
=
1.
.
.
.
(
v
)
Subtracting (iv) from (iii), we get
8
g
+
16
f
+
64
=
0
⇒
g
+
2
f
=
−
8.
.
.
.
(
v
i
)
Solving (v) and (vi), we get
g
=
−
2
and
f
=
−
3.
Putting
g
=
−
2
and
f
=
−
3
in (ii), we get
c
=
−
12.
Putting
g
=
−
2
,
f
=
−
3
and
c
=
−
12
in (i), we get
x
2
+
y
2
−
4
x
−
6
y
−
12
=
0
,
which is the required equation of the circle.
Centre of this circle =
(
−
g
,
−
f
)
=
(
2
,
3
)
.
And, its radius =
√
g
2
+
f
2
−
c
=
√
4
+
9
+
12
=
√
25
=
5
units.
I hope this answer is helpful for you
