If the centre of the circle passing through the origin is $$\left( {3,\,4} \right),$$  then the intercepts cut off by the circle on $$x$$-axis and $$y$$-axis respectively are :

Solution :
Equation of circle having radius $$r$$ and centre $$\left( {3,\,4} \right)$$  is $$ = {\left( {x - 3} \right)^2} + {\left( {y - 4} \right)^2} = {r^2}$$
if it is passing through $$\left( {0,\,0} \right)$$
$$\eqalign{ & \therefore \,{\left( {0 - 3} \right)^2} + {\left( {0 - 4} \right)^2} = {r^2} \cr & \Rightarrow {r^2} = 25 \cr} $$
Equation of circle is $${\left( {x - 3} \right)^2} + {\left( {y - 4} \right)^2} = 25$$
Putting $$y = 0$$
$$\therefore \,x = 6$$   units $$=$$ interception $$x$$-axis intercept on $$y$$ axis (putting $$x = 0$$  ) is $$y = 8$$  units.