Question

how to find radius of circumcircle if quadrilateral vertices are given ?

Answer 1

If the quadrilateral ABCD is a cyclic quadrilateral, the vertices A, B, C and D are on the circumcircle.

So the circle is also the circumcircle of triangle ABC.  Hence, the perpendicular bisectors of AB and BC intersect at the center O.

Find the slope of AB.  FInd the mid-point E of AB, it is easy as both vertices are given.  The write the equation of OE as slope of OE = -1/slope of AB.

Similarly, find the slope of BC, as both points are given.  Find the midpoint F.  Then write the equation of FO as the slope of FO = -1/slope of BC.

Now find the point of intersection of  FO and EO.  We get coordinates of O.

Find OA= OB = OC = OD = radius.

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Find the radius of a circum circle by finding the sides, a, b, and c as we know the coordinates of A, B and C.

$s= semi-perimeter=(a+b+c)/2\\\\ Radius = \frac{a*b*c}{4\sqrt{s(s-a)(s-b)(s-c)}}$

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You could do the same with triangle  BCD or ACD or ABD.

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Another way to do this directly using formulas:

  Translate the coordinate axes to the point A(x₁, y₁). So vertex A becomes A' (0, 0).
  Vertex B becomes B' (x₂', y₂') = (x₂-x₁,  y₂-y₁)
  Vertex C becomes C' (x₃', y₃') = (x₃-x₁, y₃-y₁).
  Let circumcenter be O' (Xc', Yc')

     Calculate      Denom = 2 (x₂' y₃' - x₃' y₂') .    Now find the circumcenter.

$X_c' = y_3' (x_2'^2 + y_2'^2) - y_2' (x_3'^2 + y_3'^2) / Denom\\Y_c' = X_2' (x_3'^2 + y_3'^2) - y_2' (x_2'^2 + y_2'^2) / Denom$
 
  Now, obtain    O (Xc, Yc)  by    Xc = Xc' + x1    and    Yc = Yc' + y1.
        Find distance between  O and any vertex to find radius.

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let a, b, c, and d be the sides of the cyclic quadrilateral. 
Let s = semi-perimeter = (a+b+c+d)/2.  Then, radius of circumcircle is:

$R =\frac{1}{4}* \sqrt{ \frac{(ac+bd)(ad+bc)(ab+cd)}{(s-a)(s-b)(s-c)(s-d)} }$

These are different ways of finding the radius.