Q.Equation of a circle which passes through $(3,6)$ and touches the axes is (A) $x^2+y^2+6x+6y+3=0$ (B) $x^2+y^2-6x-6y-9=0$ (C) $x^2+y^2-6x-6y+9=0$ (D) $x^2+y^2-6x+6y-3=0$

Since $(3,6)$ has both coordinates positive and the circle touches both axes, its centre is of the form $(a,a)$ with radius $a$ (first quadrant). Substituting the circle passing through $(3,6)$: $(3-a)^2+(6-a)^2=a^2$. Expanding: $9-6a+a^2+36-12a+a^2=a^2$, i.e. $a^2-18a+45=0$. By the quadratic formula, $a=\dfrac{18\pm\sqrt{324-180}}{2}=\dfrac{18\pm12}{2}$, giving $a=15$ or $a=3$. Taking $a=3$ (which is the value matching one of the printed options): $(x-3)^2+(y-3)^2=9$, expanding to $x^2+y^2-6x-6y+9=0$.

A circle touching both axes has centre $(a,a)$, radius $a$; substitute the given point and solve the resulting quadratic in $a$.

• Forgetting that touching both axes forces the centre's two coordinates to be equal (to the radius)
• The quadratic in $a$ has two valid roots geometrically (two different circles through the point, touching both axes) — only one matches the printed options