c1:x2+y2=r2;‌ center ‌=(0,0)‌ and radius ‌=r c2:(x−1)2+(y−1)2=r2;‌ center ‌=(1,1)‌ and radius ‌=r c3:(x−2)2+(y−1)2=r2;‌ center ‌=(2,1)‌ and radius ‌=r Distance of y=mx+c line from center (0,0) is, |‌
0+0+c
√m2+1
‌
‌
|=r.....(1) Distance of y=mx+c line from center (1,1) is, |‌
m−1+c
√m2+1
‌
‌
|=r....(2) Distance of y=mx+c line from center (2,1) is, |‌
2m−1+c
√m2+1
‌
‌
|=r....‌ (3) ‌ From (1) and (2), we get |‌
c
√1+m2
‌
‌
|=|‌
m−1+c
√1+m2
‌
‌
| ⇒m−1+c=±c taking positive sign, m−1+c=c ⇒m−1=0 From (2) and (3), we get |‌
m−1+c
√1+m2
‌
‌
|=|‌
2m−1+c
√m2+1
‌
‌
| ⇒(m−1+c)=±(2m−1+c) taking positive sign, m−1+c=2m−1+c ⇒m=0 By taking positive sign we get two different value of m so it is not acceptable. From equation (4), taking negative sign, m−1+c=−c ⇒m+2c−1=0 From equation (5), taking negative sign m−1+c=−(2m−1+c) ⇒3m+2c−2=0....(7) Solving equation (6) and (7), we get 3m+1−m−2=0