Question

Equation of the circle through origin which cuts intercepts of length a and b on axes is

• A. $x^2+y^2+ax+by=0$
• B. $x^2+y^2-ax-by=0$
  
• C. $x^2+y^2+bx+ay=0$
• D. none of these

Answer 1

The correct option is B

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

Centre of the circle is $(\frac{a}{2},\frac{b}{2})$ and its radius
is $\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{b}{2}\right)}^2}=\frac{1}{2}\sqrt{a^2+b^2}$

Equation of circle:

${(x-\frac{a}{2})}^2+{(y-\frac{b}{2})}^2=\frac{1}{4}(a^2+b^2)$

$\Rightarrow (2x-a{)}^2+(2y-b{)}^2=(a^2+b^2)$

$\Rightarrow 4x^2+a^2-4ax+4y^2+b^2-4by=a^2+b^2$

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