Question

Equation of a circle passing through the origin and making intercept by the line $4x+3y=12$ with coordinate axes is

• A. $x^2+y^2+3x+4y=0$
• B. $x^2+y^2+3x-4y=0$
• C. $x^2+y^2-3x+4y=0$
• D. $x^2+y^2-3x-4y=0$

Answer 1

The correct option is D $x^2+y^2-3x-4y=0$

The coordinates of points $A$ and $B$ are $(3,0)$ and $(0,4)$ respectively.

Let the equation of circle is

$x^2+y^2+2gx+2fy+c=0$ .....(i)

Since, this circle passes through $(0,0),(3,0)$ and

$(0,4)$ respectively, then

$c=0$ ......(ii)

$9+6g+c=0$ ......(iii)

and $16+8f+c=0$ ......(iv)

On solving equations (i), (ii) and (iii), we get

$9+6g=0\Rightarrow g=-\frac{9}{6}=-\frac{3}{2}$

$16+8f=0\Rightarrow f=\frac{-16}{8}=-2$

$\therefore c=0,g=-\frac{3}{2}$ and $f=-2$

On putting these value in equation (i), we get

$x^2+y^2-3x-4y=0$

which is required equation of circle.