Matrices & DeterminantsmediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Eigenvalue p=8 and Circle Axis Intersections: 3 Points | JEE 2026

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

Question
Let A=[1274−2838−7]A=\begin{bmatrix}1&2&7\\4&-2&8\\3&8&-7\end{bmatrix} and det⁡(A−αI)=0\det(A-\alpha I)=0, where α\alpha is a real number. If the largest possible value of α\alpha is pp, then the circle (x−p)2+(y−2p)2=320(x-p)^2+(y-2p)^2=320 intersects the coordinate axes at
A11 point
B22 points
C33 pointscorrect
D44 points
Solution
Step 1: Expand det⁡(A−αI)=0\det(A-\alpha I)=0:
(1−α)[(α+2)(α+7)−64]−2[−28−4α−24]+7[32+6+3α]=0.(1-\alpha)\big[(\alpha+2)(\alpha+7)-64\big]-2\big[-28-4\alpha-24\big]+7\big[32+6+3\alpha\big]=0.
 This simplifies to
α3+8α2−88α−320=0.\alpha^3+8\alpha^2-88\alpha-320=0.
 Step 2: Factor:
(α−8)(α2+16α+40)=0 ⇒ α=8 or α=−8±26.(\alpha-8)(\alpha^2+16\alpha+40)=0\ \Rightarrow\ \alpha=8\ \text{or}\ \alpha=-8\pm2\sqrt6.
 The largest value is α=8\alpha=8, so p=8p=8 (and 2p=162p=16). [Booklet note: the printed line "So p = -8" is a typo; the very next step uses centre (8,16)(8,16), i.e. p=8p=8, since 82+162=3208^2+16^2=320.] Step 3: Circle: (x−8)2+(y−16)2=320(x-8)^2+(y-16)^2=320. Step 4: Put y=0y=0: (x−8)2+256=320⇒(x−8)2=64⇒x=16,0(x-8)^2+256=320\Rightarrow(x-8)^2=64\Rightarrow x=16,0. Put x=0x=0: 64+(y−16)2=320⇒(y−16)2=256⇒y=32,064+(y-16)^2=320\Rightarrow(y-16)^2=256\Rightarrow y=32,0. Step 5: The intersection points are (16,0),(0,32),(0,0)(16,0),(0,32),(0,0) — the point (0,0)(0,0) is shared, giving 33 distinct points. Correct answer: (3)
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