AP EAPCET 23-Aug-2021 Shift 2 Paper

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.