The number of circles passes through (3,-6) and touches both the axes are
The correct option is B 0
Equation of circle touches both the axes is given by,
$$(x-a)^2+(y-a)^2 = a^2$$
Given it also passes through $$(3,-6)$$
$$\Rightarrow (3-a)^2+(a+6)^2=a^2\Rightarrow a^2+6a+45=0$$
Clearly this quadratic does not posses real roots.Hence no circle is possible.