Question

Equation of circle which touches the coordinate axes and passes through the point 2,1 is

Answer 1

Equation of the circle that touches the coordinate axes is given by,

$(x-a)^2+(y-a)^2=a^2$

Since this circle passes through $(2,1)$, we need

$(2-a)^2+(1-a)^2=a^2\\ 4-4a+a^2+1-2a+a^2=a^2\\ a^2-6a+5=0\\ (a-5)(a-1)=0\\ a=5,a=1$

The possible equations of the circles are

$(x-1)^2+(y-1)^2=1\\ (x-5)^2+(y-5)^2=25$