CircleshardPYQ · JEE Main · 6 Apr 2026 · Shift 1 (Morning)Free

Chord Length Squared on x=1: 80 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let the centre of the circle x2+y2+2gx+2fy+25=0x^2+y^2+2gx+2fy+25=0 be in the first quadrant and lie on the line 2x−y=42x-y=4. Let the area of an equilateral triangle inscribed in the circle be 27327\sqrt3. Then the square of the length of the chord of the circle on the line x=1x=1 is
Solution
Answer: 80 (± 0.01)
Step 1: Centre (−g,−f)(-g,-f) on 2x−y=42x-y=4:
2(−g)−(−f)=4⇒−2g+f=4⇒f=4+2g.2(-g)-(-f)=4\Rightarrow-2g+f=4\Rightarrow f=4+2g.
 Step 2: Inscribed equilateral triangle: side s=r3s=r\sqrt3, area 34s2=334r2\dfrac{\sqrt3}{4}s^2=\dfrac{3\sqrt3}{4}r^2:
334r2=273⇒r2=27⋅43=36.\dfrac{3\sqrt3}{4}r^2=27\sqrt3\Rightarrow r^2=27\cdot\dfrac{4}{3}=36.
 Step 3: r2=g2+f2−25=36⇒g2+f2=61r^2=g^2+f^2-25=36\Rightarrow g^2+f^2=61. Substitute f=4+2gf=4+2g:
g2+(4+2g)2=61⇒g2+16+16g+4g2=61⇒5g2+16g−45=0.g^2+(4+2g)^2=61\Rightarrow g^2+16+16g+4g^2=61\Rightarrow5g^2+16g-45=0.
g=−16±256+90010=−16±3410⇒g=95 or −5.g=\dfrac{-16\pm\sqrt{256+900}}{10}=\dfrac{-16\pm34}{10}\Rightarrow g=\dfrac95\ \text{or}\ -5.
 g=−5g=-5: centre (5,−(4−10))=(5,6)(5,-(4-10))=(5,6) ∴ first quadrant, accepted. g=95g=\tfrac95: centre (−95,−385)\left(-\tfrac95,-\tfrac{38}{5}\right) ∴ rejected. So centre (5,6), r=6(5,6),\ r=6. Step 4: x=1x=1 into (x−5)2+(y−6)2=36(x-5)^2+(y-6)^2=36:
16+(y−6)2=36⇒(y−6)2=20⇒y−6=±20.16+(y-6)^2=36\Rightarrow(y-6)^2=20\Rightarrow y-6=\pm\sqrt{20}.
 Chord length =220=2\sqrt{20}.
∴(chord)2=(220)2=80.\therefore(\text{chord})^2=(2\sqrt{20})^2=80.
 Correct answer: 80
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