JEE Main 25 Feb 2021 Shift 2 Solved Paper

Given, circle ⇒(x−3)2+y2=9
Parabola ⇒y2=4x
Let equation of common tangent be‌y=mx+‌

a

m

‌y=mx+‌

1

m

⇒m2x−my+1=0
∴ Perpendicular from (3,0) to line =3
⇒‌‌|‌

(3m2−0+1)

√m2+m4

|=3
⇒‌‌‌‌(3m2+1)2=9(m2+m4)⇒m=±‌

1

√3

Tangent is y=‌

1

√3

x+√3
⇒‌‌m=‌

1

√3

or ‌‌y=‌

(−1)

√3

×(−√3)‌‌ (rejected)
For parabola, point of contact is (c,d)=(‌

a

m2

,‌

2a

n

)
∴‌‌(c,d)=(3,2√3)
Again, solve circle and line equation, we get
(x−3)2+[(‌

1

√3

x+√3)]2‌‌=9
⇒‌‌x2+9−6x+‌

1

3

x2+3+2x‌‌=9
‌

4

3

x2−4x+3‌‌=0
⇒‌‌x‌‌=‌

3

2

=a
∴‌‌2(a+c)‌‌=2(‌

3

2

+3)=9