Question

A circle passes through origin and meets the axes at $A$ and $B$ so that $(2,3)$ lies on $\overline{AB}$ then the of centerid of $△OAB$ is

• A. $2x-3y=6xy$
• B. $2x+3y=6xy$
• C. $3x-2y=3xy$
• D. $3x+2y=3xy$

Answer 1

The correct option is D $3x+2y=3xy$

Circle: $x^2+y^2+2gx+2fy+c=0$

$\because$ It passes through origin.

$\Rightarrow c=0$

$\Rightarrow x^2+y^2+2gx+2fy=0$

When $y=0$,

$\Rightarrow x^2+2gx=0$

$\Rightarrow x(x+2g)=0$

$\Rightarrow x=0,-2g$

$\therefore A\to (-2g,0)$

When $x=0$,

$y^2+2fy=0$

$\Rightarrow y(y+2f)=0$

$\Rightarrow y=0,-2f$

$\therefore B\to (0,-2f)$

Equation of $AB$:

$\frac{-x}{2g}-\frac{y}{2f}=1$

$(2,3)$ lies on this line.

$\Rightarrow -\frac{2}{2g}-\frac{3}{2f}=1$

$\Rightarrow \frac{1}{g}+\frac{3}{2f}=-1\to \left(1\right)$

Controid of $OAB$ :

$(\frac{0-2g+0}{3},\frac{0+0-2f}{3})$

$\Rightarrow (\frac{-2g}{3},\frac{-2f}{3})$

$\Rightarrow$ Let $\frac{-2g}{3}=x$

$\Rightarrow g=\frac{-3x}{2}$

Let $\frac{-2f}{3}=y$

$\Rightarrow f=\frac{-3y}{2}$

Putting in $\left(1\right)$

$\Rightarrow \frac{-2}{3x}+\frac{3}{-3y}=-1$

$\Rightarrow \frac{-2}{3x}-\frac{3}{3y}=-1$

$\frac{2y+3y}{3xy}=1$

$\Rightarrow 2y+3x=3xy$

$\therefore D)$ Answer.