JEE Main 30 June 2022 Shift 1 Solved Paper

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

⇒2m=1
⇒m=‌

1

2

∴2c=1−‌

1

2

⇒c=‌

1

4

Putting value of m=‌

1

2

and c=‌

1

4

in equation (1), we get
r=|‌

‌ 1 4

√1+‌ 1 4

‌

‌

|
=|‌

1

4

×‌

2

√5

‌

‌

|
=‌

1

2√5

∴20(r2+c)
=20(‌

1

4×5

+‌

1

4

)
=20(‌

1+5

20

)
=6