Question

A circle which passes through origin and cuts intercepts on axes $a$ and $b$ . the equation of circle is

• A. $x^2+y^2-ax-by=0$
• B. $x^2+y^2+ax+by=0$
• C. $x^2+y^2-ax+by=0$
• D. $x^2+y^2+ax-by=0$

Answer 1

The correct option is A $x^2+y^2-ax-by=0$

$\therefore radius=\sqrt{\left({\frac{a}{2}}^2\right)+\left({\frac{b}{2}}^2\right)}=\left(\frac{1}{2}\right)\sqrt{a^2+b^2}$

$\therefore equationofcircle(x-(\frac{a}{2}){)}^2+(y-\left(\frac{b}{2}\right){)}^2=\left(\right(\frac{1}{2}\sqrt{}{a}^2+b^2\left)\right)$

$x^2-ax+(\frac{a^2}{4}+y^2-by+(\frac{b^2}{4}\left)\right)$

$=(\frac{a^2}{4}+(\frac{b^2}{4}\left)\right)$

$\therefore x^2+y^2-ax-by=0$