JEE 2022 Mathematics PYQ — Let the tangent to the circle at the point intersect the circle ,… | Mathem Solvex | Mathem Solvex
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JEE 2022 — Mathematics PYQ
JEE | Mathematics | 2022
Let the tangent to the circle C1:x2+y2=2 at the point M(−1,1) intersect the circle C2:(x−3)2+(y−2)2=5, at two distinct points A and B. If the tangents to C2 at the points A and B intersect at N, then the area of the triangle ANB is equal to:
Choose the correct answer:
A.
21
B.
32
C.
61
(Correct Answer)
D.
35
Correct Answer:
61
Explanation
Solution
1. Equation of the Tangent to C1
The equation of the tangent to the circle x2+y2=a2 at the point (x1,y1) is xx1+yy1=a2.
For C1:x2+y2=2 at M(−1,1):
x(−1)+y(1)=2⟹−x+y=2⟹x−y+2=0
This line L:x−y+2=0 acts as a chord of contact for circle C2 from point N(h,k).
2. Chord of Contact for C2
Let the coordinates of point N be (h,k). The equation of the chord of contact for circle (x−x0)2+(y−y0)2=r2 from external point (h,k) is:
(x−x0)(h−x0)+(y−y0)(k−y0)=r2
For C2:(x−3)2+(y−2)2=5:
(x−3)(h−3)+(y−2)(k−2)=5
Expanding this:
x(h−3)−3h+9+y(k−2)−2k+4=5
x(h−3)+y(k−2)−3h−2k+8=0
3. Finding Point N
Comparing the line x−y+2=0 with the chord of contact equation:
1h−3=−1k−2=2−3h−2k+8
From the first two parts: h−3=−k+2⟹h+k=5
From the first and third: 2h−6=−3h−2k+8⟹5h+2k=14
Solving these equations: h=34,k=311. So, N=(34,311).
4. Area of Triangle ANB
The distance p from the center of C2(3,2) to the line x−y+2=0:
p=12+(−1)2∣3−2+2∣=23
The radius of C2 is R=5.
Length of chord AB=2R2−p2=25−29=221=2.
Distance from N to the line AB:
d=2∣34−311+2∣=2∣−37+36∣=321
Area=21×Base(AB)×Height(d)
Area=21×2×321=61
Correct Option: (C) 61
Explanation
Solution
1. Equation of the Tangent to C1
The equation of the tangent to the circle x2+y2=a2 at the point (x1,y1) is xx1+yy1=a2.
For C1:x2+y2=2 at M(−1,1):
x(−1)+y(1)=2⟹−x+y=2⟹x−y+2=0
This line L:x−y+2=0 acts as a chord of contact for circle C2 from point N(h,k).
2. Chord of Contact for C2
Let the coordinates of point N be (h,k). The equation of the chord of contact for circle (x−x0)2+(y−y0)2=r2 from external point (h,k) is:
(x−x0)(h−x0)+(y−y0)(k−y0)=r2
For C2:(x−3)2+(y−2)2=5:
(x−3)(h−3)+(y−2)(k−2)=5
Expanding this:
x(h−3)−3h+9+y(k−2)−2k+4=5
x(h−3)+y(k−2)−3h−2k+8=0
3. Finding Point N
Comparing the line x−y+2=0 with the chord of contact equation:
1h−3=−1k−2=2−3h−2k+8
From the first two parts: h−3=−k+2⟹h+k=5
From the first and third: 2h−6=−3h−2k+8⟹5h+2k=14
Solving these equations: h=34,k=311. So, N=(34,311).
4. Area of Triangle ANB
The distance p from the center of C2(3,2) to the line x−y+2=0:
