the equation of the circle passing through the points of intersection of the two orthogonal circles S1 = x^2+y^2+kx-4y-1=0 S2=3x^2+3y^2-14x+23y-15=0and passing through the point (-1,-1) is
Answers
Answer:
the equation of any curve through the points of intersection of the circles (1) and (2) will be
(x
2
+y
2
+13x−3y)+k(2x
2
+2y
2
+4x−7y−25)=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+13−3)+k(2+2+4−7−25)=0
12−24k=0
k=
2
1
Now, Subsitute the value of k in an first equation, we get
(x
2
+y
2
+13x−3y)+
2
1
(2x
2
+2y
2
+4x−7y−25)=0
2x
2
+2y
2
+15x−
2
13y
−
2
25
=0
4x
2
+4y
2
+30x−13y−25=0
Answer:
Step-by-step explanation:
Correct option is
A
4x
2
+4y
2
+30x−13y−25=0
the equation of any curve through the points of intersection of the circles (1) and (2) will be
(x
2
+y
2
+13x−3y)+k(2x
2
+2y
2
+4x−7y−25)=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+13−3)+k(2+2+4−7−25)=0
12−24k=0
k=
2
1
Now, Subsitute the value of k in an first equation, we get
(x
2
+y
2
+13x−3y)+
2
1
(2x
2
+2y
2
+4x−7y−25)=0
2x
2
+2y
2
+15x−
2
13y
−
2
25
=0
4x
2
+4y
2
+30x−13y−25=0