The equation of a circle whose center is on the parabola x ^ 2 = 2Y and is tangent to the Quasilinear and Y-axis of the parabola

The equation of a circle whose center is on the parabola x ^ 2 = 2Y and is tangent to the Quasilinear and Y-axis of the parabola

The center O (x 0, y 0) of the center O (x 0, y 0) to the Y-axis and the distances to y = -1 / 2 are all equal. That is: the square of the two sides of the square of the two sides of R. the square of the two sides of the square of the two sides, we get: x0 \35\;; (Y0 + 1 / 2) = (Y0 + 1 / 2) / \\\\\\35\178;, and because o is on the parabparabparabola, x0, x0 \\\\\\\\\\\\\\\\\it's 17

The equation of a circle whose center is on the parabola y2 = 2x (Y > 0) and tangent to the Quasilinear and x-axis of the parabola is ()
A. x2+y2−x−2y−14＝0B. x2+y2+x-2y+1=0C. x2+y2-x-2y+1=0D. x2+y2−x−2y+14＝0

Let the coordinates of the center of the circle be (B22, b), then B22 + 12 = B & nbsp; can be obtained from the tangent of the circle to the Quasilinear and x-axis of the parabola; so B = 1, so the center of the circle is (12, 1) radius r = 1, so the equation of the circle whose center is on the parabola y2 = 2x (Y > 0) and tangent to the Quasilinear and x-axis of the parabola is (x − 12) 2 + (Y − 1) 2 & nbsp; = 1, that is, X2 + Y2 − x − 2Y + 14 = 0, so D is selected

The equation for finding the circle whose center is at the focus of the parabola y = x ^ 2 / 16 and tangent to the straight line 5x + 2y-4 = 0

The focus of parabola y = x ^ 2 / 16 is (0,4),
The distance from it to the known straight line is | 8-4 | / √ (25 + 4) = 4 / √ 29,
Therefore, the equation of the circle is x ^ 2 + (y-4) ^ 2 = 16 / 29