SOLUTION: Hai Sir, please help me on this problem, i got stuck 2)The centre of a circle is (2x-1,7) and it passes through the point (-3,-1).If the diameter of the circle is 20units ,then

 
The letter x is normally a variable, not an unknown constant.  However the x
in (2x-1,7) represents an unknown constant.  So to avoid conflict of notation
involving x as a variable and x as an unknown constant, I will change the x
in the point to "a", which is a standard letter to use for an unknown constant.
So let's pretend the problem was stated this way instead:
 

  
The equation of a circle with center (h,k) and radius r is
 
(x - h)² + (y - k)² = r²
 
So we substitute (h,k) = (2a-1,7), and since the diameter
of the circle is 20 units, the radius is one-half that or r=10 
 
(x - (2a-1))² + (y - 7)² = 10²
 

Since we know that (-3,-1) is a point on the circle, we can substitute
(x,y) = (-3,-1)
 

(-3 - (2a-1) )² + (-1 - 7)² = 10²
 
     (-3 - 2a + 1)² + (-8)² = 100
 
              (-2-2a)² + 64 = 100
 
                   (-2-2a)² = 36
 
                    -2 - 2a = ±√36
 
                    -2 - 2a = ±6

Using the +          Using the -

-2 - 2a =  6         -2 - 2a = -6                        
    -2a =  8             -2a = -4
      a = -4               a = 2

(x - (2a-1))² + (y - 7)² = 10²

Using the + that becomes                 

(x - (2(-4)-1))² + (y - 7)² = 10²   

(x - (-8-1))² + (y - 7)² = 10²

(x - (-9))² + (y - 7)² = 10²

(x + 9)² + (y - 7)² = 10²

That's this red circle:



-----------------------


(x - (2a-1))² + (y - 7)² = 10²

Using the - that becomes                 

(x - (2(2)-1))² + (y - 7)² = 10²   

(x - (4-1))² + (y - 7)² = 10²

(x - (3))² + (y - 7)² = 10²

(x - 3)² + (y - 7)² = 10²

That's the green circle:



Notice they both go through the point (-3,-1); in fact they intersect
at that point and both have radius 10 and diameter 20. 

So the answers are a = -4 and a = 2.  

Edwin