Given the first circle:
x²+ y² − 4x + 6y + 4 = 0
x²+ y² − 4x + 6y = - 4
x² − 4x + 4 + y² + 6y + 9 = - 4 + 4 + 9
(x - 2)² + (y + 3)² = 9
The center of the circle is the point (2, -3) and the radius r = 3
Now, for the second circle:
x² + y² + 6x + 4y + 9 = 0
x² + 6x + 9 + y² + 4y + 4 = - 9 + 9 + 4
(x + 3)² + (y + 2)² = 4.
The center of the circle is the point (- 3, - 2) and the radius r = 2
Finally, we need to find the equation of the line ( y = mx + b), that crosses through the points (- 3, - 2) and (2, -3):
[tex]\begin{gathered} \text{m = }\frac{y_2-y_1}{x_2-x_1} \\ m\text{ = }\frac{-3\text{ + 2}}{2\text{ + 3}}\text{ = }\frac{-1}{5} \end{gathered}[/tex]To find b, we replace one point on the linear equation:
y = mx + b
-3 = (-1/5)(2) + b
b = -13/5
So, the equation of the line is:
y = (-1/5)x - 13/5