Circle equation

Question

Definition of problem:

Write the circle equation which touches the coordinate axis and cross the point $M(2,1).$

I'm confused because I'm used to solve problems with given center but in this problem I cant figure out the formula.

The other problems use the following formula: $$(x-p)^2+(y-q)^2=r^2.$$

So I need a bit help, just to know the way this problem is taking.

Answer 1

Given that the circle touches the coordinate axes so ... It should be having coinciding roots at x=0 and y=0 respectively , now second degree eqn for circle is x^2 + y^2 + 2gx +2fy + c= 0 centered at (-g,-f) which is clear and c is any constant here the radius is given by sqrt(g^2+f^2-c), now put the coinciding root condition and you will get g^2=f^2=c(|g| = |f|) , so either g=-f or g=f , out of which g=f only holds true.. , substitute this as well as x=2,y=1 into general equation to make it an equation in one variable either g or f and solve to get coordinates of the center values of g and f and hence the radius finally giving you the answer..

Answer 2

For a circle to tangent both axis in the first quadrant, the center must be at $(r,r)$.

Expressing that the circle is through $(2,1)$:

$$(2-r)^2+(1-r)^2=r^2,$$or $$r^2-6r+5=0=(r-1)(r-5).$$