JEE 2023 Mathematics PYQ — Let the centre of a circle be and its radius . Let and be two tan… | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
Let the centre of a circle C be (α,β) and its radius r < 8. Let 3x+4y=24 and 3x−4y=32 be two tangents and 4x+3y=1 be a normal to C. Then, (α−β+r) is equal to:
Choose the correct answer:
A.
5
B.
6
C.
7
(Correct Answer)
D.
9
Correct Answer:
7
Explanation
1. Normal ki property ka upyog:
Hume pata hai ki kisi bhi circle ki normal hamesha uske centre se guzarti hai. Isliye, centre (α,β) equation 4x+3y=1 ko satisfy karega:
4α+3β=1⟹3β=1−4α⟹β=31−4α…(i)
2. Tangents aur Radius ka sambandh:
Centre (α,β) se tangents par perpendicular distance radius r ke barabar hota hai.
Tangents hain: L1:3x+4y−24=0 aur L2:3x−4y−32=0.
Distance formula d=a2+b2∣ax1+by1+c∣ ka use karte hue:
r=5∣3α+4β−24∣…(ii)
r=5∣3α−4β−32∣…(iii)
3. Equations ko solve karna:
Eq (ii) aur (iii) se hume milta hai:
∣3α+4β−24∣=∣3α−4β−32∣
Do cases bante hain:
Case 1:3α+4β−24=3α−4β−32⟹8β=−8⟹β=−1.
Eq (i) mein β=−1 rakhne par: −3=1−4α⟹4α=4⟹α=1.
Case 2:3α+4β−24=−(3α−4β−32)⟹3α+4β−24=−3α+4β+32⟹6α=56⟹α=28/3. (Isse check karne par r > 8 aayega, jo question ke against hai).
4. Radius r nikalna:
α=1,β=−1 ko Eq (ii) mein rakhein:
r=5∣3(1)+4(−1)−24∣=5∣3−4−24∣=5∣−25∣=5
Yahan r=5, jo ki r < 8 ki condition ko satisfy karta hai.
5. Final Answer:
Hume (α−β+r) ki value nikalni hai:
α−β+r=1−(−1)+5=1+1+5=7
Explanation
1. Normal ki property ka upyog:
Hume pata hai ki kisi bhi circle ki normal hamesha uske centre se guzarti hai. Isliye, centre (α,β) equation 4x+3y=1 ko satisfy karega:
4α+3β=1⟹3β=1−4α⟹β=31−4α…(i)
2. Tangents aur Radius ka sambandh:
Centre (α,β) se tangents par perpendicular distance radius r ke barabar hota hai.
Tangents hain: L1:3x+4y−24=0 aur L2:3x−4y−32=0.
Distance formula d=a2+b2∣ax1+by1+c∣ ka use karte hue:
r=5∣3α+4β−24∣…(ii)
r=5∣3α−4β−32∣…(iii)
3. Equations ko solve karna:
Eq (ii) aur (iii) se hume milta hai:
∣3α+4β−24∣=∣3α−4β−32∣
Do cases bante hain:
Case 1:3α+4β−24=3α−4β−32⟹8β=−8⟹β=−1.
Eq (i) mein β=−1 rakhne par: −3=1−4α⟹4α=4⟹α=1.
Case 2:3α+4β−24=−(3α−4β−32)⟹3α+4β−24=−3α+4β+32⟹6α=56⟹α=28/3. (Isse check karne par r > 8 aayega, jo question ke against hai).
4. Radius r nikalna:
α=1,β=−1 ko Eq (ii) mein rakhein:
r=5∣3(1)+4(−1)−24∣=5∣3−4−24∣=5∣−25∣=5
Yahan r=5, jo ki r < 8 ki condition ko satisfy karta hai.
