Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
If the circle x2 + y2-17x + 2fy + c = 0 passes through (3,1), (14,-1) and (11,5), then c is
Q. If the circle
x
2
+
y
2
−
17
x
+
2
f
y
+
c
=
0
passes through
(
3
,
1
)
,
(
14
,
−
1
)
and
(
11
,
5
)
, then
c
is
7029
205
Conic Sections
Report Error
A
0
23%
B
−
41
30%
C
−
17/2
18%
D
41
29%
Solution:
Given equation of circle is
x
2
+
y
2
−
17
x
+
2
f
y
+
c
=
0
.
Since it passes through
(
3
,
1
)
,
(
14
,
−
1
)
and
(
11
,
5
)
∴
9
+
1
−
51
+
2
f
+
c
=
0
or
2
f
+
c
=
41
...
(
i
)
and
196
+
1
−
238
−
2
f
+
c
=
0
or
−
2
f
+
c
=
41
...
(
ii
)
Adding
(
i
)
and
(
ii
)
, we get
2
c
=
(
2
×
41
)
or
c
=
41
.