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 $\Delta$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

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

$B(3,0)$ & $C(-3,0)$

A lie on the circle with BC as diameter center of circle = midpoint of BC=$(0,0)$

Radius of circle $=\frac{BC}{2}=3$

$\therefore$ locus of A is $x_1^2+y_1^2=9$

Centroid of $BAC=(\frac{x_1+3-3}{3},\frac{y_1+0+0}{3})$

$=(\frac{x_1}{3},\frac{y_1}{3})$

$(x,y)=(\frac{x_1}{3},\frac{y_1}{3})$

$\Rightarrow x_1=3x$ and $y_1=3y$

$\therefore$ Locus of centroid is

$x_1^2+y_1^2=9$

$\Rightarrow (3x{)}^2+(3y{)}^2=9$

$\Rightarrow x^2+y^2=1$

$\therefore$ Locus of centroid of $△ABC$ is

$x^2+y^2=1$