Solve
Study
Textbooks
Guides
Use app
Login
Question
The equation of a circle passing through the point
(
1
,
1
)
and the point of intersection of the circles
x
2
+
y
2
+
1
3
x
−
3
y
=
0
and
2
x
2
+
2
y
2
+
4
x
−
7
y
−
2
5
=
0
is
A
4
x
2
+
4
y
2
+
3
0
x
−
1
3
y
−
2
5
=
0
B
4
x
2
+
4
y
2
+
3
0
x
−
1
3
y
+
2
5
=
0
C
4
x
2
−
4
y
2
−
3
0
x
+
1
3
y
−
2
5
=
0
D
4
x
2
−
4
y
2
+
3
0
x
−
1
3
y
−
2
5
=
0
Hard
Open in App
Solution
Verified by Toppr
Correct option is A)
the equation of any curve through the points of intersection of the circles (1) and (2) will be
(
x
2
+
y
2
+
1
3
x
−
3
y
)
+
k
(
2
x
2
+
2
y
2
+
4
x
−
7
y
−
2
5
)
=
0
It is given that equation of circles passes through the point
(
1
,
1
)
So,
x
=
1
and
y
=
1
Substitute these value in an above equation, we get
(
1
+
1
+
1
3
−
3
)
+
k
(
2
+
2
+
4
−
7
−
2
5
)
=
0
1
2
−
2
4
k
=
0
k
=
2
1
Now, Subsitute the value of k in an first equation, we get
(
x
2
+
y
2
+
1
3
x
−
3
y
)
+
2
1
(
2
x
2
+
2
y
2
+
4
x
−
7
y
−
2
5
)
=
0
2
x
2
+
2
y
2
+
1
5
x
−
2
1
3
y
−
2
2
5
=
0
4
x
2
+
4
y
2
+
3
0
x
−
1
3
y
−
2
5
=
0
Video Explanation
Was this answer helpful?
0
0
Similar questions
Find the equation of circle passing through the points of intersection of
x
2
+
y
2
=
6
and
x
2
+
y
2
−
6
x
+
8
=
0
and the point
(
1
,
1
)
Medium
View solution
>
Let ABCD be a square such that vertices A,B,C,D lie on circles
x
2
+
y
2
−
2
x
−
2
y
+
1
=
0
,
x
2
+
y
2
+
2
x
−
2
y
+
1
=
0
,
x
2
+
y
2
+
2
x
+
2
y
+
1
=
0
and
x
2
+
y
2
−
2
x
+
2
y
+
1
=
0
respectively with centre of square being origin and sides are parallel to coordinate axes. The length of the side of such a square can be
This question has multiple correct options
Hard
View solution
>
The equation of the circle passing through the point of intersection of the circle
x
2
+
y
2
+
2
x
+
3
y
+
1
=
0
,
x
2
+
y
2
+
4
x
+
3
y
+
2
=
0
and through the point
(
−
1
,
2
)
is
Easy
View solution
>
If a circle passes through the point (a,b) and cuts the circle
x
2
+
y
2
=
p
2
orthogonally , then the equation of the locus of its centre is
Medium
View solution
>
The centre of the circle passing through the point
(
0
,
1
)
and touching the curve
y
=
x
2
at
(
2
,
4
)
is
Hard
View solution
>