1. Given data

We need the circles which:

• pass through the point $P(-9,4)$,
• touch both lines $x+y=3 \, , \quad x-y=3.$

We must find the absolute difference of the squares of their radii.

2. Observation about centers of circles touching both lines

If a circle touches both intersecting lines, then its center lies on one of the angle bisectors of the two lines.

The lines are: $x+y-3=0, \quad x-y-3=0.$

Their angle bisectors satisfy $\frac{x+y-3}{\sqrt{1^2+1^2}}=\pm \frac{x-y-3}{\sqrt{1^2+(-1)^2}}.$

Since denominators are same, $x+y-3=\pm (x-y-3).$

This gives:

• for $+$: $x+y-3=x-y-3 \Rightarrow y=-y \Rightarrow y=0$,
• for $-$: $x+y-3=-x+y+3 \Rightarrow 2x=6 \Rightarrow x=3$.

So the center lies on either $y=0 \quad \text{or} \quad x=3.$

3. Case 1: center on $y=0$

Let the center be $C(h,0).$

Since the radius equals perpendicular distance from the center to either tangent line, $r=\frac{|h+0-3|}{\sqrt 2}=\frac{|h-3|}{\sqrt 2}.$

Also the circle passes through $P(-9,4)$, so $CP=r.$

Thus, $\sqrt{(h+9)^2+(0-4)^2}=\frac{|h-3|}{\sqrt 2}.$

Squaring, $(h+9)^2+16=\frac{(h-3)^2}{2}.$

Multiply by 2: $2(h+9)^2+32=(h-3)^2.$

Expand: $2(h^2+18h+81)+32=h^2-6h+9.$ $2h^2+36h+162+32=h^2-6h+9.$ $h^2+42h+185=0.$

Discriminant: $42^2-4\cdot 185=1764-740=1024=32^2.$

So, $h=\frac{-42\pm 32}{2}=-21\pm 16.$

Hence $h=-5,\,-37.$

Now radius squares: $r^2=\frac{(h-3)^2}{2}.$

• For $h=-5$: $r_1^2=\frac{(-8)^2}{2}=32.$
• For $h=-37$: $r_2^2=\frac{(-40)^2}{2}=800.$

Their difference is $|800-32|=768.$

4. Case 2: center on $x=3$

Let the center be $C(3,k).$

Then radius is the distance to either line: $r=\frac{|3+k-3|}{\sqrt2}=\frac{|k|}{\sqrt2}.$

Since the circle passes through $(-9,4)$, $\sqrt{(3+9)^2+(k-4)^2}=\frac{|k|}{\sqrt2}.$

Squaring, $144+(k-4)^2=\frac{k^2}{2}.$

Multiply by 2: $288+2(k-4)^2=k^2.$

Expand: $288+2(k^2-8k+16)=k^2,$ $288+2k^2-16k+32=k^2,$ $k^2-16k+320=0.$

Discriminant: $(-16)^2-4\cdot 320=256-1280=-1024<0.$

So there is no real circle from this case.

5. Conclusion

Thus the only two circles are from Case 1, with $r_1^2=32, \qquad r_2^2=800.$

Hence the required absolute difference is $|r_2^2-r_1^2|=|800-32|=768.$

6. Comparison with stored answer

Stored correct answer = $768$.

Our derived answer also is $768$, so they agree.