The intersection of any two diameters of a circle gives the center of the circle.\n\nWe solve the given equations of the diameters to find the center \( (h, k) \):\n\[ 2x + 3y = 1 \quad \text{--- (1)} \]\n\[ 4x - 3y = 11 \quad \text{--- (2)} \]\n\nAdding equations (1) and (2):\n\[ 6x = 12 \implies x = 2 \]\n\nSubstitute \( x = 2 \) into (1):\n\[ 2(2) + 3y = 1 \implies 4 + 3y = 1 \implies 3y = -3 \implies y = -1 \]\nThe center of the circle is \( (2, -1) \).\n\nGiven the area of the circle is 153.94:\n\[ \pi r^2 = 153.94 \]\nUsing \( \pi \approx \frac{22}{7} \) or 3.14:\n\[ r^2 = \frac{153.94}{3.14} \approx 49 \]\nSince the radius must be an integer, \( r = 7 \).\n\nThe equation of the circle is:\n\[ (x - h)^2 + (y - k)^2 = r^2 \]\n\[ (x - 2)^2 + (y + 1)^2 = 7^2 \]\n\[ x^2 - 4x + 4 + y^2 + 2y + 1 = 49 \]\n\[ x^2 + y^2 - 4x + 2y - 44 = 0 \]