Question

Equation of a circle which passes through the origin and cuts off intercepts $\frac{2}{3}$ and $\frac{3}{2}$ respectively from the axes of $x$ and $y$ is

• A. $6x^2+6y^2+4x+9y=0$
• B. $6x^2+6y^2-4x+9y=0$
• C. $6x^2+6y^2+4x-9y=0$
• D. $6x^2+6y^2-4x-9y=0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $6x^2+6y^2-4x+9y=0$
B $6x^2+6y^2+4x+9y=0$
C $6x^2+6y^2+4x-9y=0$
D $6x^2+6y^2-4x-9y=0$

Given that, circle passes through origin, and cut of intercepts $\frac{2}{3}$ and $\frac{3}{2}$ respectively on $x$ and $y$ axes.
Let $x^2+y^2+2gx+2fy+c=0$ be the equation of the required circle.
$c=0$ as it passes through origin.
x-intercept $2\sqrt{g^2-c}=\frac{2}{3}$
$\Rightarrow g=\pm \frac{1}{3}$
y-intercept $2\sqrt{f^2-c}=\frac{3}{2}$
$\Rightarrow f=\pm \frac{3}{4}$
Required equation of circle is $x^2+y^2\pm \frac{2}{3}x\pm \frac{3}{2}y=0$
$\therefore$ Circle equation is $6x^2+6y^2\pm 4x\pm 9y=0$
Hence, options A,B,C and D.