Consider the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-10\mathrm{x}-6\mathrm{y}+30=0$. Let O be the center of the circle and tangents at A(7 , 3) and
B(5, 1) me et at C. Let S = 0 represents the family of circles passing through A and B. Then

The coordinates of O are (5, 3) and the radius is 2.
The equation of tangent at A(7 ,3) is
$7\mathrm{x}+3\mathrm{y}-5(\mathrm{x}+7)-3(\mathrm{y}+3)+30=0$
i.e., 2x - 14 = 0
i.e., x = 7
The equation of tangent at B(5, 1) is
$5\mathrm{x}+\mathrm{y}-5(\mathrm{x}+5)-3(\mathrm{y}+1)+30=0$
i.e., $-2\mathrm{y}+2=0$
i.e., y = 1
Therefore, the coordinates of C are (7,1). So,
area of OACB = 4
The equation of AB is x - y = 4 (radical axis).
The equation of the smallest circle is
$(\mathrm{x}-7)(\mathrm{x}-5)+(\mathrm{y}-3)(\mathrm{y}-1)=0$
i.e., ${\mathrm{x}}^2+{\mathrm{y}}^2-12\mathrm{x}-4\mathrm{y}+38=0$