Find equation of largest circle passing through $(1,1)$ and $(2,2)$ and which does not cross boundaries of first quadrant. 
Find equation of largest circle passing through $(1,1)$ and $(2,2)$ and which does not cross boundaries of first quadrant.

my attempt
I tried writing a family through line $x=y$ and $(1,1)$ and $(2,2)$ as
$$(x-1)(x-2)+(y-1)(y-2)+p(x-y)=0$$
where $p$ is a variable parameter.
I tried to make $g^2<c$ and $f^2<c$; so as to have no $x,y$ intercepts. But I am having trouble maximising the radius. I was looking for any suggestions or any better solutions.

The  circles passing through $A$ and $B$ are a linear combination of the equation of the line $AB:x-y=0$ and the circle having diameter $AB$, $\mathscr{C}:x^2+y^2-3x-3y+4=0$
Therefore its general equation is $x^2+y^2-3x-3y+4+k(x-y)=0$
$x^2+y^2-3x-3y+4+k(x-y)=0$ intersect $x-$axis if $y=0$. Substitute in the circle and get
$x^2-(3-k)x+4=0$ 
This equation has non real roots if discriminant is less than zero 
$\Delta=(3-k)^2-16<0 \to k^2-6k-7<0\to -1<k<7$
The circle intersects the $y-$axis if $x=0$ and with the same procedure as above we get no intersection when $k^2+6 k-7<0$ that is $-7<k<1$
So the condition for $k$ to have circles non intersecting axes is $-1<k<1$
Therefore the maximum radius is achieved when $k=\pm 1$
Collect $x$ and $y$ in the previous equation we get
$x^2+y^2+(k-3) x-(k+3) y+4=0$
radius is $r=\sqrt{\left(\dfrac{k-3}{2}\right)^2+\left(\dfrac{k+3}{2}\right)^2-4}$
for $k=\pm 1$ we have $r=1$ which is the maximum radius 
Requested circles have equations
$x^2+y^2-4 x-2 y+4=0$
$x^2+y^2-2 x-4 y+4=0$
Hope this helps

