JEE 2023 Mathematics PYQ — Let . Let be the circle of radius in the first quadrant touching … | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
Let ω=zzˉ+k1z+k2iz+λ(1+i),k1,k2∈R. Let Re(ω)=0 be the circle C of radius 1 in the first quadrant touching the line y=1 and the y-axis. If the curve Im(ω)=0 intersects C at A and B, then 30(AB)2 is equal to ________.
Choose the correct answer:
A.
24
(Correct Answer)
B.
25
C.
26
D.
27
Correct Answer:
24
Explanation
Step 1: ω ko simplify karna
Sabse pehle z=x+iy rakhte hain:
ω=(x+iy)(x−iy)+k1(x+iy)+k2i(x+iy)+λ(1+i)
ω=(x2+y2)+k1x+ik1y+k2ix−k2y+λ+iλ
Ab Real aur Imaginary parts ko alag karte hain:
Re(ω)=x2+y2+k1x−k2y+λ
Im(ω)=k1y+k2x+λ
Step 2: Circle C ki equation (Re(ω)=0)
Humein diya gaya hai ki Re(ω)=0 ek circle hai jo first quadrant mein hai, radius 1 hai, y-axis ko touch karta hai aur line y=1 ko bhi touch karta hai.
Agar radius 1 hai aur circle y-axis ko touch karta hai, toh center ka x-coordinate 1 hoga.
Agar radius 1 hai aur circle y=1 (horizontal line) ko touch karta hai, toh center ka y-coordinate 1+radius=2 hoga (kyunki circle line ke upar ya niche ho sakta hai, par "first quadrant" aur "radius 1" ke hisaab se center (1,2) ya (1,0) ho sakta tha. Par y=1 ko touch karne ke liye y-coordinate 0 ya 2 hona chahiye. First quadrant mein center (1,2) hi fit baithta hai).
Circle ki equation:(x−1)2+(y−2)2=12
x2−2x+1+y2−4y+4=1⟹x2+y2−2x−4y+4=0
Re(ω) se compare karne par:
k1=−2,k2=4,λ=4
Step 3: Curve Im(ω)=0 (Line)
Ab k1,k2,λ ki values Im(ω) mein rakhte hain:
−2y+4x+4=0⟹4x−2y+4=0⟹2x−y+2=0
Yani line ki equation hai: y=2x+2
Step 4: Chord AB ki length nikaalna
Line 2x−y+2=0 aur circle jiska center O(1,2) aur radius r=1 hai, unke intersection points A aur B hain. Chord length ka formula hota hai:
AB=2r2−p2
Jahan p center (1,2) se line 2x−y+2=0 ki perpendicular distance hai.
p=22+(−1)2∣2(1)−(2)+2∣=5∣2∣=52
Ab AB nikaalte hain:
AB=212−(52)2=21−54=251=52
Step 5: Final Calculation
Humein 30(AB)2 ki value chahiye:
(AB)2=(52)2=54
30(AB)2=30×54=6×4=24
Final Answer:
24
Explanation
Step 1: ω ko simplify karna
Sabse pehle z=x+iy rakhte hain:
ω=(x+iy)(x−iy)+k1(x+iy)+k2i(x+iy)+λ(1+i)
ω=(x2+y2)+k1x+ik1y+k2ix−k2y+λ+iλ
Ab Real aur Imaginary parts ko alag karte hain:
Re(ω)=x2+y2+k1x−k2y+λ
Im(ω)=k1y+k2x+λ
Step 2: Circle C ki equation (Re(ω)=0)
Humein diya gaya hai ki Re(ω)=0 ek circle hai jo first quadrant mein hai, radius 1 hai, y-axis ko touch karta hai aur line y=1 ko bhi touch karta hai.
Agar radius 1 hai aur circle y-axis ko touch karta hai, toh center ka x-coordinate 1 hoga.
Agar radius 1 hai aur circle y=1 (horizontal line) ko touch karta hai, toh center ka y-coordinate 1+radius=2 hoga (kyunki circle line ke upar ya niche ho sakta hai, par "first quadrant" aur "radius 1" ke hisaab se center (1,2) ya (1,0) ho sakta tha. Par y=1 ko touch karne ke liye y-coordinate 0 ya 2 hona chahiye. First quadrant mein center (1,2) hi fit baithta hai).
Circle ki equation:(x−1)2+(y−2)2=12
x2−2x+1+y2−4y+4=1⟹x2+y2−2x−4y+4=0
Re(ω) se compare karne par:
k1=−2,k2=4,λ=4
Step 3: Curve Im(ω)=0 (Line)
Ab k1,k2,λ ki values Im(ω) mein rakhte hain:
−2y+4x+4=0⟹4x−2y+4=0⟹2x−y+2=0
Yani line ki equation hai: y=2x+2
Step 4: Chord AB ki length nikaalna
Line 2x−y+2=0 aur circle jiska center O(1,2) aur radius r=1 hai, unke intersection points A aur B hain. Chord length ka formula hota hai:
AB=2r2−p2
Jahan p center (1,2) se line 2x−y+2=0 ki perpendicular distance hai.
