27123335−2−(λ+4)=0⇒12(−21)−3(−39)−(λ+4)(−15)=0⇒−252+117+15(1+4)=0⇒15λ+177−252=0⇒15λ−75=0⇒λ=59816−μ3335−2−9=0⇒1μ−816−μ00377−9=0⇒7−7(μ−8)=0⇒1−(μ−8)=0⇒μ=9⇒ centre of circle (5,9) radius = length of ⊥ from centre (5,9)=520−27=57
JEE Main 2025 — Mathematics Algebra
Let the system of equations :
$\begin{aligned}
& 2 x+3 y+5 z=9 \
& 7 x+3 y-2 z=8 \
& 12 x+3 y-(4+\lambda) z=16-\mu
\end{aligned}$
have infinitely many solutions. Then the radius of the circle centred at (λ,μ) and touching the line 4x=3y is
Held on 7 Apr 2025 · Verified 6 Jul 2026.
517
57
7
521
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let the set of all values of $k \in \mathbb{R}$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0$, $z \in \mathbb{C}$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:
If $\alpha=1$ and $\beta=1+i\sqrt{2}$, where $i=\sqrt{-1}$ are two roots of the equation $x^3+ax^2+bx+c=0$, $a,b,c \in \mathbb{R}$, then $\int_{-1}^{1}(x^3+ax^2+bx+c)dx$ is equal to:
Let $\alpha = 3+4+8+9+13+14+\ldots$ upto 40 terms. If $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2+x-2=0$, $\beta \in \left(0, \dfrac{\pi}{2}\right)$, then $\sin^2\beta + 3\cos^2\beta$ is equal to:
$\frac{6}{3^{26}}+\frac{10 \cdot 1}{3^{25}}+\frac{10 \cdot 2}{3^{24}}+\frac{10 \cdot 2^{2}}{3^{23}}+\ldots+\frac{10 \cdot 2^{24}}{3}$ is equal to :
Let $\mathrm{C}_{\mathrm{r}}$ denote the coefficient of $x^{\mathrm{r}}$ in the binomial expansion of $(1+x)^{\mathrm{n}}, \mathrm{n} \in \mathrm{N}, 0 \leq \mathrm{r} \leq \mathrm{n}$. If $P_{n}=C_{0}-C_{1}+\frac{2^{2}}{3} C_{2}-\frac{2^{3}}{4} C_{3}+\ldots. .+\frac{(-2)^{n}}{n+1} C_{n}$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2 n}}$ equals.
Work through every JEE Main Algebra PYQ, year by year.