The absolute difference between the squares of the radii of the two circles passing through the point $$(-9, 4)$$ and touching the lines $$x + y = 3$$ and $$x - y = 3$$, is equal to _____.

Let the two given lines be $$L_1 : x + y = 3$$ and $$L_2 : x - y = 3$$.

Step 1 - Find their intersection and show they are perpendicular.
Solving the two equations simultaneously gives the point of intersection $$I(3,0)$$.
The slopes are $$-1$$ for $$L_1$$ and $$+1$$ for $$L_2$$ and their product is $$-1$$, hence the lines are perpendicular.

Step 2 - Locate the possible centres.
For a circle to be tangent to two perpendicular lines, its centre must lie on one of the two angle-bisectors of the lines.
The angle-bisectors of the pair $$Y = \pm X$$ (after shifting the origin to $$I(3,0)$$ by putting $$X = x-3,\; Y = y$$) are
  • $$Y = 0 \; \Longrightarrow\; y = 0$$, a horizontal line through $$I$$, and
  • $$X = 0 \; \Longrightarrow\; x = 3$$, a vertical line through $$I$$.

We therefore examine two cases.

Case 1: Centre $$C(h,0)$$ lies on $$y = 0$$.

Radius as distance from the centre to either line:
$$r = \dfrac{|h-3|}{\sqrt2} \; \Longrightarrow\; r^{2} = \dfrac{(h-3)^{2}}{2}$$.

Radius as distance from the centre to the fixed point $$P(-9,4)$$:
$$r^{2} = (h+9)^{2} + (0-4)^{2} = (h+9)^{2} + 16$$.

Equating the two expressions for $$r^{2}$$ gives
$$\dfrac{(h-3)^{2}}{2} = (h+9)^{2} + 16$$.
Multiplying by $$2$$ and simplifying:
$$(h-3)^{2} = 2(h+9)^{2} + 32$$
$$h^{2} - 6h + 9 = 2h^{2} + 36h + 194$$
$$0 = h^{2} + 42h + 185$$.

Solving the quadratic:
$$h = \dfrac{-42 \pm \sqrt{42^{2} - 4\cdot1\cdot185}}{2} = \dfrac{-42 \pm 32}{2}$$
$$\Longrightarrow\; h = -5 \quad\text{or}\quad h = -37$$.

Thus two centres are $$C_1(-5,0)$$ and $$C_2(-37,0)$$.
Corresponding radii:
$$r_1^{2} = ( -5 + 9)^{2} + 16 = 4^{2} + 16 = 32 \;\; \Longrightarrow\;\; r_1 = 4\sqrt2$$,
$$r_2^{2} = ( -37 + 9)^{2} + 16 = 28^{2} + 16 = 800 \;\; \Longrightarrow\;\; r_2 = 20\sqrt2$$.

Case 2: Centre $$C(3,k)$$ lies on $$x = 3$$.

Radius from the centre to either line:
$$r = \dfrac{|k|}{\sqrt2} \; \Longrightarrow\; r^{2} = \dfrac{k^{2}}{2}$$.

Radius from the centre to the fixed point $$P(-9,4)$$:
$$r^{2} = (3 + 9)^{2} + (k - 4)^{2} = 144 + (k - 4)^{2}$$.

Equating gives
$$\dfrac{k^{2}}{2} = 144 + (k - 4)^{2}$$
$$k^{2} = 288 + 2(k^{2} - 8k + 16)$$
$$k^{2} = 2k^{2} - 16k + 320$$
$$0 = k^{2} - 16k + 320$$.

The discriminant of this quadratic is
$$\Delta = (-16)^{2} - 4\cdot1\cdot320 = 256 - 1280 = -1024 \lt 0,$$
so there is no real value of $$k$$. Consequently, no circle has its centre on $$x = 3$$.

Step 3 - Compute the required absolute difference.
The only admissible radii are $$r_1 = 4\sqrt2$$ and $$r_2 = 20\sqrt2$$.
Therefore
$$|\,r_2^{2} - r_1^{2}\,| = |\,800 - 32\,| = 768.$$

Hence, the absolute difference between the squares of the radii is 768.