Two circles of radius 4 are tangent to the graph of y^2=4x at the point (1,2). Find the equation of these two circles.

Consider the line tangent to y^2=4x. It has m = x^(-1/2). At 1,2, that's m=1, so y=x+1. The circles are tangent to this line, so a perpendicular through (1,2) joins their centres. That'll be y=-x+3.

So to get our centres we're looking for two points with a distance of 4 from (1,2) along y=-x+3.

Since the slope is -1, the x-distance must be the same as the y-distance, so the distance in both directions must be root(2).

Given that, you have your centres, and plugging that and the radius into the standard equation of a circle finishes the job.

(x-(1-(4/sqrt(2))))^2+(y-(2+(4/sqrt(2))))^2=16

and

(x-(1+(4/sqrt(2))))^2+(y-(2-(4/sqrt(2))))^2=16

To find the equation of the circles, we need to determine their centers.

Given that the circles are tangent to the graph of y^2=4x at the point (1,2), we can infer that the centers of the circles lie on the line y=2.

Now, let's identify the x-coordinates of the centers. Since the circles are tangent to the graph, the distance from the center to the point (1,2) should be equal to the radius.

The distance from (1,2) to any point on the line y=2 is given by the equation: √((x-1)^2 + (y-2)^2).

Since the radius is 4, we have the following equation:

√((x-1)^2 + (2-2)^2) = 4

Simplifying, we get:

√((x-1)^2) = 4

(x-1)^2 = 16

x-1 = ±4

x = 1 ± 4

Hence, the x-coordinates of the centers of the two circles are x=5 and x=-3.

Since the circles are tangent to the graph y^2=4x, the x-coordinate of the point of tangency should be equidistant from the centers.

For the circle with center (5,2), let's determine the equation of the circle using the standard form equation: (x-h)^2 + (y-k)^2 = r^2.

Center = (5,2)
Radius = 4

Equation of the first circle:

(x-5)^2 + (y-2)^2 = 16

For the circle with center (-3,2), we have:

Center = (-3,2)
Radius = 4

Equation of the second circle:

(x+3)^2 + (y-2)^2 = 16

Therefore, the equations of the two circles tangent to the graph y^2=4x at the point (1,2) are:

(x-5)^2 + (y-2)^2 = 16

and

(x+3)^2 + (y-2)^2 = 16

To find the equation of the circles, we need to determine their centers.

Let's start by finding the equation of the tangent line to the graph of y^2 = 4x at the point (1, 2). Since the graph is symmetric with respect to the y-axis, the slope of the tangent line at (1, 2) will be 0.

The equation of the tangent line can be written as y = mx + b, where m is the slope and b is the y-intercept. Since the slope is 0, the equation simplifies to y = b.

Substituting the coordinates of the point (1, 2) into the equation, we get:
2 = b

So the equation of the tangent line is y = 2.

Since the two circles are tangent to this line, their centers must lie on the line y = 2.

Now, let's find the x-coordinate of each center. Since the circles have a radius of 4, their centers will have a distance of 4 units to the right and left of the y-axis. Therefore, the x-coordinates of the centers will be 4 and -4.

Using the y-coordinate of 2 and the x-coordinate of either 4 or -4, we can write the equations of the circles as follows:

Center: (4, 2)
Equation of circle: (x - 4)^2 + (y - 2)^2 = 4^2
Simplifying, we get: (x - 4)^2 + (y - 2)^2 = 16

Center: (-4, 2)
Equation of circle: (x + 4)^2 + (y - 2)^2 = 4^2
Simplifying, we get: (x + 4)^2 + (y - 2)^2 = 16

Therefore, the equations of the two circles are:
(x - 4)^2 + (y - 2)^2 = 16
and
(x + 4)^2 + (y - 2)^2 = 16