questionr
Early access
Exams
Joint Entrance Examination
Mathematics
Circle
Question 196
Question 196 of 219
Medium
MCQ
Choose the correct statement about two circles whose equations are given below : x2 + y2 - 10x - 10y + 41 = 0 x2 + y2 - 22x - 10y + 137 = 0
A
circles have same centre
B
circles have no meeting point
C
circles have only one meeting point
D
circles have two meeting points
Check answer
Press A-D to select · Enter to check
Solutions · 1
Post your solution
Official solution
Let
S
1
:
x
2
+
y
2
−
10
x
−
10
y
+
41
=
0
{S_1}:{x^2} + {y^2} - 10x - 10y + 41 = 0
S
1
:
x
2
+
y
2
−
10
x
−
10
y
+
41
=
0
\[ \Rightarrow {(x - 5)^2} + {(y - 5)^2} = 9\] Centre
(
C
1
)
=
(
5
,
5
)
({C_1}) = (5,5)
(
C
1
)
=
(
5
,
5
)
Radius r1 = 3 \[{S_2}:{x^2} + {y^2} - 22x - 10y + 137 = 0\] \[ \Rightarrow {(x - 11)^2} + {(y - 5)^2} = 9\] Centre
(
C
2
)
=
(
11
,
5
)
({C_2}) = (11,5)
(
C
2
)
=
(
11
,
5
)
Radius r2 = 3 distance \[({C_1}{C_2}) = \sqrt {{{(5 - 11)}^2} + {{(5 - 5)}^2}} \] distance
(
C
1
C
2
)
=
6
({C_1}{C_2}) = 6
(
C
1
C
2
)
=
6
\[ \because \]
r
1
+
r
2
=
3
+
3
=
6
{r_1} + {r_2} = 3 + 3 = 6
r
1
+
r
2
=
3
+
3
=
6
\[ \therefore \] circles touch externally Hence, circle have only one meeting point.
Answer the question to unlock
Previous
Next