Give the center and radius of the points B(1,4) C(-1,2) A(4, -3) My procedure:
First, I solved for the slope of line CA. Afterwards, I solved for its Midpoint. Then I formed its equation. 
I also applied the same process to line CB. According to my research, I should subtract the first equation to the second equation. I'm confused because the value of my x was a fraction after subtracting. I decided to make it zero. Then, I used it and solved for the value of y. I got 25/4. I solved for the midpoint by the points C and (0,25/4) I got the center (2,13/5)  I guess converse of angle in a semicircle is involved with this.

I'm probably wrong with my calculation, so I'm open for any corrections. Please explain your answer in detailed. Thank you.
If you are comfortable solving three linear equations in three unknowns, let the equation of the circle be $$(x-h)^2+(y-k)^2 = r^2.$$ Substitute points $A$, $B$ and $C$, you will obtain three linear equations with unknowns $h$, $k$ and $(r^2-h^2-k^2)$.
Here are two different approaches:

This one seems more direct to me.

There exists a point $X = (x,y)$ such that  $d(X,A) = d(X,B) = d(X,C)$

$(x-4)^2 + (y+3)^2  = (x-1)^2 + (y-4)^2 = (x+1)^2 + (y-2)^2\\
x^2 - 8x + 16 + y^2 + 6x + 9 = x^2 - 2x + 1 + y^2 - 8y + 16 = x^2 + 2x + 1 + y^2 - 4y + 4$ 

We can subtract $x^2 + y^2$ across the board. 

$ - 8x +  6x + 25 =  - 2x  - 8y + 17 =  2x  - 4y +5$

and consolidate a little bit further

$ - 10x +  10x + 20 =  - 4x  - 4y + 12 = 0$

and break this up into two equations.

$ - 10x +  10x + 20 = 0\\
 - 4x  - 4y + 12 = 0$

And solve for $x$ and $y.$

To find the radius.

$r^2 = (x-4)^2 + (y+3)^2  = (x-1)^2 + (y-4)^2 = (x+1)^2 + (y-2)^2$

plug the value of $x,y$ into any of the above.

The approach you suggest works.

Find the lines through the midpoint of $AC$ and $BC,$ with slopes perpendicular to $AC$ and $BC$

Slope of $AC = -1$

Slope of the perpendicular $= 1$

mid-point $= (1.5, -0.5)$

$y+0.5 = x - 1.5\\
-x + y = -2$

Slope of $BC = 1$

Slope of the perpendicular $=-1$

mid-point $= (0, 3)$

$y-3 = - x\\
 x + y = 3$

and again, a system of 2 linear equations.

And you will need to apply a distance formula to find the radius.