The equation of circle touching the parabola at upper end of its latus rectum and passing through its vertex, is

The problem seems to be incomplete as it does not provide the equation of the parabola. However, I will provide a general solution for a parabola y^2 = 4ax.

Step 1: Identify the vertex and the end of the latus rectum of the parabola. The vertex of the parabola y^2 = 4ax is at the origin (0,0). The latus rectum of this parabola is the line segment parallel to the x-axis passing through the focus (a,0). The length of the latus rectum is 4a, so its endpoints are (a, 2a) and (a, -2a). The upper end of the latus rectum is the point (a, 2a).

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

Step 3: Substitute the known points into the circle equation. The circle passes through the vertex of the parabola (0,0) and touches the parabola at the upper end of its latus rectum (a, 2a). Substituting these points into the circle equation gives two equations:

For (0,0): h^2 + k^2 = r^2 For (a, 2a): (a-h)^2 + (2a-k)^2 = r^2

Step 4: Solve the system of equations. Subtract the first equation from the second to eliminate r^2 and solve for h and k. The solution will give the center of the circle. Substitute h and k back into the first equation to find r^2, the square of the radius.

Step 5: Write down the equation of the circle. Substitute h, k, and r into the general equation of the circle to get the equation of the circle that touches the parabola at the upper end of its latus rectum and passes through its vertex.