Obviously, the required circle is the circumcircle of the triangle ABC. The perpendicular bisectors of AB, BC, & CA intersect at the centre of the circumcircle. Let us find the equaton of the perpendicular bisectors of AB & BC.
Mipoint of AB is ( (2+5) / 2, (2+3)/2 ) i.e (7/2, 5/2)
Slope of AB is (3-2)/(5-2) i.e 1/3
Slope of the perpendicular bisector of AB is - 3.
Equation of perpendicular bisector of AB is
y - 5/2 = - 3 ( x - 7/2) ........ eqn (1)
Mipoint of BC is ( (5+3) / 2, (3-1)/2 ) i.e (4, 1)
Slope of BC is (3+1)/(5-3) i.e 2
Slope of the perpendicular bisector of BC is - 1/2.
Equation of perpendicular bisector of AB is
y - 1 = - 1/2 ( x - 4) ........ eqn (2)
Solving (1) & (2),
we have, x = 4 and y = 1
i.e the coordinates of the centre of the circle is (4,1)
Now, the radius of the circle is = the distance of either A,B or C i.e the vertices from the centre = 
[ (4-2)2 + (1-2)2 ] = 5 units.
So, the equation of the required circle passing through A,B, C is
(x - 4)2 + (y - 1)2 = 5
Again we know any point on the circle,
( x -c)2 + (y-d)2 = r2 can be written as ( c + rcosA , d + rsinA )
Thus any point on the circle,
(x - 4)2 + (y - 1)2 = 5 can be written as (4+5 cos theta, 1+5 sin theta )
Alternative method of finding the centre of the circle
The triangle ABC is a rightt angled triangle.
[ slope of AB . slope of CA = 1/3 . -3 = -1 ]
For any right angled triangle the circumcentre is at the midpoint of the hypotenuse. As such the circumcentre of the triabgle ABC is the midpoint of BC i.e (4,1)
The latter method of finding the circumcentre is a special method while the former one can be applied for all types of triangle.
Cheers !!!
Moderation Team
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