Let $$f\left( {x,\,y} \right) = 0$$   be the equation of a circle. If $$f\left( {0,\,\lambda } \right) = 0$$   has equal roots $$\lambda = 2,\,2,$$   and $$f\left( {\lambda ,\,0} \right) = 0$$   has roots $$\lambda = \frac{4}{5},\,5$$   then the centre of the circle is :

Solution :
$$\eqalign{ & f\left( {x,\,y} \right) \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0 \cr & {\text{Now,}}\,{\text{ }}f\left( {0,\,\lambda } \right) \equiv {\lambda ^2} + 2f\lambda + c = 0,\,{\text{and its roots are }}2,\,2 \cr & \therefore \,\,2 + 2 = - 2f,\,2 \times 2 = c,\,\,{\text{i}}{\text{.e}}{\text{.,}}f = - 2,\,c = 4 \cr & f\left( {\lambda ,\,0} \right) \equiv {\lambda ^2} + 2g\lambda + c = 0,\,{\text{and its roots are }}\frac{4}{5},\,5 \cr & \therefore \,\,\frac{4}{5} + 5 = - 2g,\,\frac{4}{5} \times 5 = c \cr & {\text{Thus }}g = - \frac{{29}}{{10}},\,f = - 2 \cr} $$