Question

Find the equation of the circle which passes through the origin and the cuts off chords of the lengths $4$ and $6$ on the positive side of $x-axis$ and $y-axis$ respectively.

• A. $x^2+y^2-4x-6y=0$
• B. $x^2-y^2+4x+6y=0$
• C. $-x^2+y^2+4x-6y=0$
• D. None of these

Answer 1

The correct option is A $x^2+y^2-4x-6y=0$

general equation of circle is $x^2+y^2+2gx+2fy+c=0$

length of chord on $x-$ axis $=4$

length of chord on $y-$ axis $=6$

Observe that $(0,0),(4,0),(0,6)$ are on circle

So putting these points in the general equation we have

$0+0+2g.0+2f.0+c=0.....\left(i\right)$

$16+0+8g+2f.0+c-0.....\left(ii\right)$

$0+36+2g.0+12f+c=0.....\left(iii\right)$

$\left(i\right)$ gives $c=0$

$\left(ii\right)$ gives $g=-2$

$\left(iii\right)$ gives $f=-3$

So equation of circle is $x^2+y^2-4x-6y=0$