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Maths Question 4 – JEE-MAIN 2025

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Let the system of equations : 2x+3y+5z=9, 7x+3y−2z=8, 12x+3y−(4+λ)z=16−μ, have infinitely many solutions. Then the radius of the circle centred at (λ,μ) and touching the line 4x=3y is

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the determinants of the matrices formed by replacing a column with the constant terms must also be zero.

Step 1: Determine conditions for infinitely many solutions✦ Active

For the given system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero. The coefficient matrix is:

|23573−2123−(4+λ)|=0

Expanding the determinant, we get:

2(3(−(4+λ))−3(−2))−3(7(−(4+λ))−12(−2))+5(7(3)−12(3))=0 ⟹2(−12−3λ+6)−3(−28−7λ+24)+5(21−36)=0 ⟹2(−6−3λ)−3(−4−7λ)+5(−15)=0 ⟹−12−6λ+12+21λ−75=0 ⟹15λ−75=0⟹λ=5

For consistency with infinitely many solutions, the determinant Dz (formed by replacing the z-column with the constant terms) must also be zero:

|23973812316−μ|=0

Expanding this determinant, we get:

2(3(16−μ)−3(8))−3(7(16−μ)−12(8))+9(7(3)−12(3))=0 ⟹2(48−3μ−24)−3(112−7μ−96)+9(21−36)=0 ⟹2(24−3μ)−3(16−7μ)+9(−15)=0 ⟹48−6μ−48+21μ−135=0 ⟹15μ−135=0⟹μ=9

Thus, the center of the circle is (λ,μ)=(5,9).

Step 2: Calculate the radius of the circle○ Expand

The circle is centered at (5,9) and touches the line 4x=3y, which can be rewritten as 4x−3y+0=0. The radius R of the circle is the perpendicular distance from its center (x0,y0) to the tangent line Ax+By+C=0. The formula for this distance is:

R=|Ax0+By0+C|A2+B2

Substituting the values (x0,y0)=(5,9), A=4, B=−3, C=0:

R=|4(5)−3(9)+0|42+(−3)2 ⟹R=|20−27|16+9 ⟹R=|−7|25 ⟹R=75

The radius of the circle is 75.

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