Let center of the circle is (h,k) and the radius is r.

Since circle touches the y-axis, so x-coordinate of the center of the circle is same as the radius r of the circle.

Now the equation of the circle having center (h,k) and radius r is

(x-h)² + (y-k)² = r²

Since h = r, then

      (x-r)² + (y-k)² = r²

=> x² - 2rx + r² + y² - 2ky + k² = r²

=> x² - 2rx + y² - 2ky + k² = 0 ..............1

Agian given that the circle passes through the points (-2,1) and (-4,3)

Put (x,y) = (-2, 1) in equation 1, we get

      (-2)² - 2r*(-2) + 1² - 2k*1 + k² = 0

=> 4 + 4r + 1 - 2k + k² = 0

=> k² - 2k + 4r + 5 = 0 .................2

Again put (x,y) = (-4, 3) in equation 1, we get

      (-4)² - 2r*(-4) + 3² - 2k*3 + k² = 0

=> 16 + 8r + 9 - 6k + k² = 0

=> k² - 6k + 8r + 25 = 0 .................3

Multiply by 2 in equation 2, we get

2k² - 4k + 8r + 10 = 0 .................4

equation 4 - equation 3, we get

      k² + 2k - 15 = 0

=> (k-3)*(k+5) = 0

=> k = 3, -5

To find the value of r corresponding to k = 3, substitute in equation 2, we get

      3² - 2*3 + 4r + 5 = 0

=> 9 - 6 + 4r + 5 = 0

=> 8 + 4r = 0

=> 4r = -8

=> r = -8/4

=> r = -2

To find the value of r corresponding to k = -5, substitute in equation 2, we get

      (-5)² - 2*(-5) + 4r + 5 = 0

=> 25 + 10 + 4r + 5 = 0

=> 40 + 4r = 0

=> 4r = -40

=> r = -40/4

=> r = -10

Now put k = 3 and r = -2  in equation1, we get

      x² - 2*(-2)x + y² - 2*3*y + (3)² = 0

=> x² + 4x + y² - 6y + 9 = 0

=> x² + y² + 4x - 6y + 9 = 0

Again put k = 5 and r = -10  in equation1, we get     

      x² - 2*(-10)x + y² - 2*5*y + (-10)² = 0

=> x² + 20x + y² - 10y + 100 = 0

=> x² + y² + 20x - 10y + 100 = 0