Find the equation of a circle centre on the line y=2x+1 touching the y axis and passing through A(4,5)

a general circle is

(x-h)^2 + (y-k)^2 = r^2

since the center is on y=2x+1, k=2h+1

(x-h)^2 + (y-2h-1)^2 = r^2
Since the circle passes through (4,5),

(4-h)^2 + (5-2h-1)^2 = r^2

Since the circle is tangent to the y-axis, h=r, so

(4-r)^2 + (4-2r)^2 = r^2
r = 2 or 4

You can now check to see whether those are both possible solutions, after substituting back to get (h,k).

To find the equation of a circle with its center on a given line and passing through a given point, we need to follow a few steps:

Step 1: Determine the center of the circle.
Since the circle's center is on the line y = 2x + 1, we can assume the center has coordinates (a, 2a + 1), where a is a constant.

Step 2: Find the radius of the circle.
The circle passes through point A(4, 5). The distance between the center of the circle (a, 2a + 1) and point A(4, 5) is the radius of the circle. We can use the distance formula to find the radius:
r = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Let's calculate the distance:
r = sqrt((4 - a)^2 + (5 - (2a + 1))^2)
r = sqrt((4 - a)^2 + (4 - 2a)^2)
r = sqrt((16 - 8a + a^2) + (16 - 16a + 4a^2))
r = sqrt(20 - 24a + 5a^2)

Step 3: Write the equation of the circle.
The equation of a circle with center (a, 2a + 1) and radius r is:
(x - a)^2 + (y - (2a + 1))^2 = r^2

Substituting the value of r from Step 2, we have:
(x - a)^2 + (y - (2a + 1))^2 = (sqrt(20 - 24a + 5a^2))^2

Simplifying the equation further, we get:
(x - a)^2 + (y - 2a - 1)^2 = 20 - 24a + 5a^2

That is the equation of the circle with its center on the line y = 2x + 1, passing through A(4, 5) and touching the y-axis.