Here, we have to find the equation of the circle whose centre is at (2, - 3) and which passes through the intersection of the lines 3x + 2y = 11 and 2x + 3y = 4 First let's find the point of intersection of the lines 3x + 2y = 11 and 2x + 3y = 4 So, by solving the equations 3x + 2y = 11 and 2x + 3y = 4, we get x = 5 and y = - 2 So, the required circle passes through the point (5, - 2) Let the radius of the required circle be r As we know that, the equation of circle with centre at (h, k) and radius r units is given by: (x - h) + (y - k) = r Here, we have h = 2 and k = - 3 ⇒(x−2)2+(y+3)2=r2 ........(1) ∵ The circle passes through the point (5, - 2) So, x = 5 and y = - 2 will satisfy the equation (1) ⇒(5−2)2+(−2+3)2=r2 ⇒r2=10 So, the equation of the required circle is (x−2)2+(y+3)2=10 ⇒x2+y2−4x+6y+3=0 So, the equation of the required circle is x2+y2−4x+6y+3=0 Hence, option A is the correct answer.