The equa on of circle having the centre on the x-axis is x2 + y2 + 8x + 7 = 0,
then the centre is
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Answer:
The equation of a circle passing through the point of intersection of the given circles is
x
2
+y
2
−8x−2y+7+4(x
2
+y
2
−4x+10y+15)=0
Where h is as arbitary constant
or (1+k)x
2
+(1+k)y
2
−4x(2+k)−2y(1−5k)+7+8k=0
or x
2
+y
2
−4(
1+k
2+k
)x−2(
1+k
1−5k
)y+
1+k
7+8k
=0
where k
=−1
The coordinates of center of circle is {−2
(1+k)
(2+k)
−
(1+k)
(1−5k)
}
Since center of the circle lies on y− axis.
Abscissa =0
−2(
1+k
2+k
)=0⇒k=−2
From (1) equation of required circle
x
2
+y
2
−8x−2y+2(x
2
+y
2
−4x+10y+8)=0
x
2
+y
2
+22y+9=0
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