Question

$B$ and $C$ are fixed points having co-ordinates $(3,0)$ and $(-3,0)$ respectively. If the vertical angle $BAC$ is ${90}^o$, 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. $9(x^2+y^2)=4$

Answer 1

Answer: (A)

$x^2+y^2=1$

$BAC={90}^0\Rightarrow \left(\frac{b}{a+3}\right)\left(\frac{b}{a-3}\right)=-1$
$b^2=-(a^2-9)\Rightarrow a^2+b^2=9$ ....(1)
Now $BG:GD=2:1$
$3h=\frac{2(a+3)}{2}+1\times -3\Rightarrow a=3h$
$3k=2\left(\frac{b}{2}\right)+1\times 0\Rightarrow b=3k$
Substitute value of $a$ and $b$ in equation (1), we get
$9h^2+9k^2=9$ $\Rightarrow x^2+y^2=1$