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.
 A: You can do it without messing with complicated equations.
This is not a solution but an approach that might be helpful.  
First, note that any circle passing through $(1,1)$ and $(2,2)$ will have centre on perpendicular bisector of line joining these two points as it is a chord of the circle, i.e. the line with slope $-1$ passing through $(1.5,1.5)$. 
You can parametrise (or not) the centre as a point on this line.
Now, use the fact the largest circle will just touch either the $x$ or $y$ axis.  $x$ and $y$ are equivalent for this question and hence wlog take it to touch $y$ axis.
Put
($x$-coordinate of centre $=$ distance between centre and $y$ axis $= $radius $=$ distance between point $(1,1)$ or $(2,2)$ and centre).   
Solving this will give the maximum possible radius
A: 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

A: Hint:
Let $A=(1,1)$, $B=(2,2)$. The centre of the circle will be on the perpendicular bisector of $[AB]$. This perpendicular bisector has equation 
$x+y=3$ since it passes through the point $(3/2,3/2)$.
Furthermore, the maximum radius will be attained when the distance of $C$ to point $A$ is equal to its distance to the $x$-axis if it is below line $(AB)$, OR when the distance of $C$ to point $A$ is equal to its distance to the $y$-axis if it is above line $(AB)$.
This means $C$ is on the parabola with focus $A$ and directrix the $x$-axis or the parabola with same focus and directrix the $y$-axis. Note they're symmetric w.r.t. the line $y=x$.
There remains to find the equations of these parabolæ and their intersections with the perpendicular bisector.
