Statement I The length of common chord of the circles x2+y2+ax+by+c=0 and x2+y2+bx+ay+c=0 equals
√(a+b)2−8c
2
Let the circles c1 be the center of x2+y2+ax+by+c=0 ...(i) and c2 be the center of x2+y2+bx+ay+c=0 ...(ii) ∴c1=(
−a
2
,
−b
2
) and c2=(
−b
2
,
−a
2
) r1=√
a2
4
+
b2
4
−c ⇒r1=
1
2
√a2+b2−4c r2=√
b2
4
+
a2
4
−c ⇒r2=
1
2
√a2+b2−4c ∵ Radius of both circles is same. c1c2=√(
−a
2
+
b
2
)2+(
−b
2
+
a
2
)2
⇒c1c2=√
(a−b)2
4
+
(a−b)2
4
⇒c1c2=
a−b
√2
Let AB be common chord and M is its mid-point. ∴C1M=
c1c2
2
=
a−b
2√2
∵Ac1M forms a right angle triangle. ∴AM=√r12−(c1M)2 =√
a2+b2−4c
4
−
a2+b2−2ab
8
=
√(a+b)2−8c
2√2
∴AB=2‌AM=
√(a+b)2−8c
√2
⇒ Statement- 1 is false. We know that if two circles intersect at two distinct points, then their radical axis is their common chord. ⇒ Statement-2 is true.