Let the equation of circle be 

x²+y²+2gx+2fy+c=0        .....(1)
where (−g,−f) is center and g²+f²−c is the radius.
Since, the circle passes through (2,3), so it satisfies eqn (1)
⇒4+9+4g+6f+c=0
⇒4g+6f+c=−13      .....(2)
Since, the circle passes through (−1,1), so it satisfies eqn (1)
⇒1+1−2g+2f+c=0
⇒−2g+2f+c=−2      .....(3)
Subtracting eqn (3) from eqn (2), we get
6g+4f=−11       ....(4)
Given center lies on the line x−3y−11=0
Since, center (−g,−f) satisfies this equation.
⇒−g+3f=11        ....(5)
Solving eqn (4) and (5), we get
g= −7​/2 ,f= 5​/2
Put this value in (2), we get
c=−14
Substituting these values in (1),
x²+y²−7x+5y−14=0
which is the equation of required circle.

A diagram used to represent all possible relations of different sets. A Venn diagram can be represented by any closed figure, whether it be a Circle or a Polygon (square, hexagon, etc.). But usually, we use circles to represent each set. 

In the above figure, we can see a Venn diagram, represented by a rectangular universal set, which has two independent sets X and Y. Therefore, X and Y are disjoint sets. The two sets X and Y are represented in the circular shape. This diagram shows that set X and set Y have no relation between each other, but they are a part of universal set.

The base equation for a circle centered at (h,k) with radius r is:

   (x - h)²+ (y - k)² = r²

If the points are concyclic, they all lie on the graph for the same circle.

Thus, all points are solutions to a single equation of the circle

WE can plug each point into the base equation, and obtain a system of equationd

FOR POINT (-2,10)

     Equation 1:  (-2 - h)² + (10 - k)² = r²

FOR POINT (1,11)

     Equation 2:  (1 - h)²+ (11 - k)² = r²

FOR POINT (6,10)

     Equation 3:  (6 - h)² + (10 - k)² = r²

FOR POINT (9,7)

     Equation 4:  (9 - h)² + (7 - k)² = r²

From Equation 1 and 2 above, since they have 2 different expressions equivalent to r2, we can construct an equation involving variables x and h.

     r² = r²   

     (-2 - h)² + (10 - k)² = (1 - h)² + (11 - k)²

     (-2 - h)(-2 - h) + (10 - k)(10 - k) = (1 - h)(1 - h) + (11 - k)(11 - k)

     (4 + 2h + 2h + h²) + (100 -10k - 10k + k²) = (1 - h - h + h²) + (121 - 11k - 11k + k²)

      h² + 4h + 4 + k² - 20k + 100 = h²-2h + 1 + k² -22k + 121

    We can subtract h and k from both sides.

      4h + 4 - 20k + 100 = -2h + 1 -22k + 122

      4h -20k + 104 = -2h - 22k + 122

    Move all h and k variables to the left ... move all constants to the right

      6H + 2K = 18

SYSTEM EQUATION 1:   6h + 2k = 18      

From equation 3 and 4 above, we go through the same process to obtain a second system equation.

       r² = r²

       (6 - h)² + (10 - k)² = (9 - h)² + (7 - k)²

       (6 - h)(6 - h) + (10 - k)(10 - k) = (9 - h)(9 - h) + (7 - k)(7 - k)

       36 - 6h - 6h + h²+ 100 - 10k - 10k + k² = 81 - 9h - 9h + h² + 49 = 7k - 7k + k²

       h² - 12h + 36 + k² - 20k + 100 = h² - 18h + 81 + k²- 14k + 49

    Simplify, and subtract h2 and k2from both sides

       -12h - 20k + 136 = -18h - 14k + 130

      Move the h and k variables to the left, and the constants to the right

        -12h + 18h - 20k + 14k = 130 - 136

        6h - 6k = -6

SYSTEM EQUATION 1:  6h - 6k = -6

      6h +2k = 18

      6h - 6k = -6

Multiply the second equation by -1, we have ... -6h + 6k = 6

Adding the 2 equations, we have

 6h + 2k = 18

  -6h + 6k =  6

          8k =24

             k = 3

Using substitution of k=3 into equation 6h + 2k = 18,
   
6h + 2(3) = 18

    6h + 6 = 18

    6h = 12

    h = 2

Now, use one of the points on the circle ... (1, 11) and (h,k) ... (2,3), to find the radius

   (1 - 2)² + (11 - 3)² = r²

   1 + 64 = r²

   65 = r²

   r = √65
Thus the equation of our circle is:
   (x - 2)² + (y - 3)² = (√65)²