y2=4(x+4) Equation of circle (x+3)2+y2=25 Passes through the point of intersection of two lines 3x−y=0 and x+λy=4 (3λ+14,3λ+112), we get $\begin{aligned}
& \lambda=-\frac{7}{6}, 1 \
& 12 \lambda_1+29 \lambda_2 \
& -14+29=15
\end{aligned}$
JEE Main 2025 — Mathematics Coordinate Geometry
The focus of the parabola y2=4x+16 is the centre of the circle C of radius 5 . If the values of λ, for which C passes through the point of intersection of the lines 3x−y=0 and x+λy=4, are λ1 and λ2,λ1<λ2, then 12λ1+29λ2 is equal to
Held on 23 Jan 2025 · Verified 6 Jul 2026.
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