2. i) Find the equation of the circle which passes through the point (0, -3) and the
intersects the circles x + y2 - 6x + 3y + 5 = 0 and x + y2 - x - 7y = 0 orthogonally
[May 2015TS, 2013)
Answers
Answer:
We know that the equation of circle passing through intersection of given circles is ⇒(x
2
+y
2
−2x−6y+6)+k(x
2
+y
2
+2x−6y)=0
⇒(1+k)x
2
+(1+k)y
2
+(−2+2k)x+(−6−6k)y+6=0
⇒(1+k)x
2
+(1+k)y
2
+2(k−1)x+2(−3−3k)y+6=0 ....(1)
⇒x
2
+y
2
+2(
k+1
k−1
)x+2(
k+1
−3−3k
)y+
k+1
6
=0
⇒x
2
+y
2
+2(
k+1
k−1
)x+2(−3)y+
k+1
6
=0
Given that this circle intersects x
2
+y
2
+4x+6y+4=0 orthogonally.
The condition or orthogonal intersection of 2 circles is 2(gg
′
+ff
′
)=c+c
′
⇒2(
k+1
k−1
×2+(−3)(3))=4+
k+1
6
⇒
k+1
2(k−1)
−9=2+
k+1
3
⇒
k+1
2(k−1)
−11=
k+1
3
⇒2(k−1)−11(k+1)=3
On solving we get, k=
9
16
Substituting this value of k in (1) we get,
⇒25x
2
+25y
2
+14x−150y+54=0
Step-by-step explanation: