Question

$B$ and $C$ are fixed points having co-ordinates $(3,0)$ and $(-3,0)$ respectively. If the vertical angle $BAC$ is ${90}^{\circ }$, then the locus of the centroid of the $△ABC$ has the equation

• A. $x^2+y^2=1$
• B. $x^2+y^2=2$
• C. $9(x^2+y^2)=1$
• D. $9(x^2+y^2)=19$

Answer 1

The correct option is A $x^2+y^2=1$

Let the centroid be (h,k)

$\therefore h=\frac{\alpha }{3},K=1\frac{3}{3}⟶\left(1\right)$

Squaring,

$h^2+k^2=\frac{{\alpha }^2+{\beta }^2}{y}\Rightarrow h^2+k^2=\frac{{\beta }^2}{y}[\therefore \alpha =0]$

As $\alpha A=90C{B}^2=A{B}^2+A{B}^2(-3-3{)}^2+(0-0{)}^2=(\alpha +3{)}^2+{\beta }^2+(\alpha -3{)}^2+{\beta }^{2}36=9+9+2{\beta }^{2}2{\beta }^2=18\Rightarrow {\beta }^2=9\therefore \beta =\pm 3\therefore h^2+k^2=\frac{(\pm 3)2}{9}=1h\to x,k\to y\Rightarrow x^2+y^2=1$