There is more than one way to do this, but let's do it the
way the tutor above said to do it:

The standard equation of a circle is

(x-h)² + (y-k)² = r²



The line connecting (2,1) amd (3,5) is a chord of the circle. 


So we can use the fact that the perpendicular bisector of a chord 
passes through the center of the circle. That's the red line below



We get the slope of the green chord using the slope formula:

m = 
m = 
m = 
m = 4

So the slope of the red perpendicular bisector of the green
chord is it's "negative reciprocal" or .

We get the midpoint of the green chord using the midpoint
formula:

Midpoint = 

Midpoint = 

Midpoint =  

Midpoint = 

So the red perpendicular bisector of the chord is the line with
slope  that goes through the point 

We use the point slope formula:

y - y1 = m(x - x1)

y - 3 = (x - 5/2)

y - 3 = x + 5/8

Multiply through by 8

8y - 24 = -2x + 5
2x + 8y = 29

So to find the center we solve the system



and get the point (x,y) = (,4)

That's the center (h,k) = (,4)

So we can draw in the circle:



So the equation of the circle, so far is

(x-()² + (y-4)² = r²

(x+)² + (y-4)² = r²

And since we know it goes through (2,1) we can
substitute that point:

(2+)² + (1-4)² = r²
(+)² + (-3)² = r²
()² + 9 = r²
 + 9 = r²
 +  = r²
 = r²

So the standard equation of the circle is

(x+)² + (y-4)² = 

To find the general equatrion of the circle,
nultiply that out:

x² + 3x +  + y² - 8y + 16 = 

Multiply through by 4

4x² + 12x + 9 + 4y² - 32y + 64 = 85

4x² + 12x + 73 + 4y² - 32y = 85

4x² + 4y² + 12x - 32y - 12 = 0

Divide through by 4

x² + y² + 3x - 8y - 3 = 0

That's the general equation of the circle.

Edwin

Here's another way to do the problem:




Suppose (x,y) is the center of the circle.  Then the distance from
(x,y) to (2,1) must equal the distance from (x,y) to (3,5), so
we use the diatance formula and set them equal:



 = sqrt( (x-3)^2 + (y-5)^2)}}}

(x-2)² + (y-1)² = (x-3)² + (y-5)²

Rearrange to get difference of squares on each side:

(x-2)² - (x-3)² = (y-5)² - (y-1)²

[(x-2)-(x-3)][(x-2)+(x-3)] = [(y-5)-(y-1)][(y-5)+(y-1)]

[x-2-x+3]{x-2+x-3] = [y-5-y+1][y-5+y-1]

[1][2x-5] = [-4][2y-6]

2x-5 = -8y + 24

2x + 8y = 29

So we have the system of equations:



Solve that and get the center, and do the rest like the
other solution.

Edwin