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:
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For problems involving two tangential circles and points of intersection, use the distance formula between the center and line to find the radius.
Let the center of the first circle be \( (a, 0) \) with radius \( r_1 \). Its equation is \( (x - a)^2 + y^2 = r_1^2 \). The distance from the circle's center to the line \( x + y = 3 \) equals the radius \( r_1 \). Using the distance formula for a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \), which is \( \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \), we establish a relationship between \( a \) and \( r_1 \). For the second circle, the line is \( x - y = 3 \). The computed result is the absolute difference of the squares of the radii: \( |r_1^2 - r_2^2| = 768 \).
