Let ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}=28,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=56$ and ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}=70$. Let $\mathrm{A}(4 \cos t, 4 \sin t), \mathrm{B}(2 \sin t,-2 \cos \mathrm{t})$ and $C\left(3 r-n, r^2-n-1\right)$ be the vertices of a triangle $A B…

Question

Let ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}=28,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=56$ and ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}=70$. Let $\mathrm{A}(4 \cos t, 4 \sin t), \mathrm{B}(2 \sin t,-2 \cos \mathrm{t})$ and $C\left(3 r-n, r^2-n-1\right)$ be the vertices of a triangle $A B C$, where $t$ is a parameter. If $(3 x-1)^2+(3 y)^2$ $=\alpha$, is the locus of the centroid of triangle ABC , then $\alpha$ equals

Aicharya · Accepted Answer

$\left.\begin{array}{l}{ }^n C_{r-1}=28 \\ { }^n C_r=56 \\ { }^n C_{r+1}=70 \\ \begin{array}{l}{ }^n C_{r-1} \\ { }^n C_r\end{array}=\frac{28}{56} \Rightarrow \frac{r}{n-r+1}=\frac{1}{2} \\ \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{56}{70} \Rightarrow \frac{r+1}{n-r}=\frac{70}{56}\end{array}\right\} n=8=3$

$h=\frac{4 \cos t+2 \sin t+1}{3} \quad k=\frac{4 \sin t-2 \cos t}{3}$
$3 h-1=4 \cos t+2 \sin t$
$3 k-1=4 \sin t-2 \cos t$
$\begin{aligned}
& (1)^2+(2)^2 \\ & (3 h-1)^2+(3 k)^2=20
\end{aligned}$
Locus of centroid: $(3 x-1)^2+(3 y)^2=20$
$\Rightarrow \alpha=20$ *

Let ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}=28,{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=56$ and ${…

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