Two circles of radius 4 are tangent to the graph of y^2=4x at the point (1,2). Find equations of these two circles.
I found the derivative of y^2=4x, but i don't know what to do next!
y^2 = 4x
.: dy/dx = 2/y
The derivative evaluated at (1,2) gives the tangential gradient.
The negative inverse of the gradient gives the gradient of the normal, and hence the equation of the normal line through (1,2).
The equation of the normal line, and the given radius (4) allows you to find the centres of the circles.
The centre points and the radius allow you to write the equations for the circles.
To find the equations of the two circles tangent to the graph of y^2=4x at the point (1,2), you first need to determine the center coordinates of these circles.
Let's assume the center coordinates of the first circle are (h, k). Since the circle is tangent to the graph at the point (1,2), the distance between the center of the circle and the point (1,2) is equal to the radius of the circle, which is 4.
Using the distance formula, you can set up the following equation:
√((1 - h)^2 + (2 - k)^2) = 4
Simplifying the equation:
(1 - h)^2 + (2 - k)^2 = 16
Expanding and rearranging the equation:
1 - 2h + h^2 + 4 - 4k + k^2 = 16
h^2 + k^2 - 2h - 4k - 11 = 0
Similarly, for the second circle, let's assume the center coordinates are (p, q). The distance between the center of the circle and the point (1,2) is also equal to the radius of the circle, which is 4. Using the same steps as before, you will obtain another equation in terms of p and q.
So now you have two equations:
h^2 + k^2 - 2h - 4k - 11 = 0 (equation for the first circle)
p^2 + q^2 - 2p - 4q - 11 = 0 (equation for the second circle)
These equations can be rearranged to obtain the equations of the two circles.