Question

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

Options

•  x² + y² + ax + by = 0

• x² + y² − ax − by = 0

• x² + y² + bx + ay = 0

• none of these

Answer 1

x² + y² − ax − by = 0

Centre of the circle is \[\left( \frac{a}{2}, \frac{b}{2} \right)\] 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:

\[\left( x - \frac{a}{2} \right)^2 + \left( y - \frac{b}{2} \right)^2 = \frac{1}{4}\left( a^2 + b^2 \right)\]

\[\Rightarrow\] \[\left( 2x - a \right)^2 + \left( 2y - b \right)^2 = \left( a^2 + b^2 \right)\]

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

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