JEE Main 29 June 2022 Shift 1 Solved Paper

Equation of tangent at point M is
T=0
⇒xx1+yy1=2
⇒−x+y=2
⇒y=x+2
Putting this value to equation of circle C2,
(x−3)2+(y−2)2=5
⇒(x−3)2+x2=5
⇒x2−6x+9+x2=5
⇒2x2−6x+4=0
⇒x2−3x+2=0
⇒(x−2)(x−1)=0
⇒x=1,2
when x=1,y=3
and when x=2,y=4
∴ Point A(1,3) and B(2,4)
Now, equation of tangent at A(1,3) on circle (x−3)2+(y−2)2=5 or x2+y2−6x−4y+8=0 is

T=0
xx1+yy1+g(x+x1)+f(y+y1)+C=0
⇒x+3y−3(x+1)−2(y+3)+8=0
⇒x+3y−3x−3−2y−6+8=0
⇒−2x+y−1=0
⇒2x−y+1=0
Similarly tangent at B(2,4) is
2x+4y−3(x+2)−2(y+4)+8=0
⇒2x+4y−3x−6−2y−8+8=0
⇒−x+2y−6=0
⇒x−2y+6=0
Solving equation (1) and (2), we get
x−2(2x+1)+6=0
⇒x−4x−2+6=0
⇒−3x+4=0
⇒x=‌

4

3

∴y=2×‌

4

3

+1=‌

11

3

∴‌ Point ‌N=(‌

4

3

,‌

11

3

)
Now area of the triangle ANB
=‌

1

2

|

x1	y1	1
x2	y2	1
x3	y3	1

|
=‌

1

2

|

1	3	1
2	4	1
‌ 4 3	‌ 11 3	1

|
=‌

1

2

[1(4−‌

11

3

)−3(2−‌

4

3

)+1(‌

22

3

−‌

16

3

)]
=‌

1

2

[‌

1

3

−‌

6

3

+‌

6

3

]
=‌

1

6