Two circles of radius 4 are tangent to the graph y^2=4x at the point (1,2). Find the equations of the two circles.
The given parabola y²=4x has the axis along the x-axis. So we will work with its inverse, y=2√x. The slope can be obtained by differentiation, dy/dx = 1/sqrt(x).
The slope of the tangent line at (1,2) is therefore 1/sqrt(1) = 1
The slope of the normal, N0, to the curve at this point is m=1/(-1) = -1
(y-y1)=m(x-x1), substituting m=-1, x1=1, y1=2 the normal is
N0 : y = -x +3
Since both required circles pass through (1,2), their centres will be on the circumference of a circle C0 with radius 4 centred on the point of tangency, where
C0 : (x-1)² + (y-2)² = 4²
The centres of the required circles will be found at the intersections of N0 and C0.
Can you take it from here?
To find the equations of the two circles, we need to determine their centers.
Let's consider the circle with center (h, k). Since it is tangent to the graph y^2 = 4x at the point (1, 2), the distance from its center to the point (1, 2) equals its radius, which is 4.
Using the formula for the distance between two points, we have:
√[(h - 1)^2 + (k - 2)^2] = 4
Simplifying this equation, we get:
(h - 1)^2 + (k - 2)^2 = 16 ------(1)
Since the other circle is also tangent to the graph y^2 = 4x at the point (1, 2), it will have the same equation as the first circle with a different center.
Now, let's find the coordinates of the other center, which we can denote as (m, n). Using the same logic as before, we have:
√[(m - 1)^2 + (n - 2)^2] = 4
Simplifying this equation, we get:
(m - 1)^2 + (n - 2)^2 = 16 ------(2)
Equation (1) and equation (2) are both equations of circles. Solving these two equations simultaneously will give us the equations of the two circles. Let's solve it:
Expanding the squared terms, we have:
h^2 - 2h + 1 + k^2 - 4k + 4 = 16 ------(1')
m^2 - 2m + 1 + n^2 - 4n + 4 = 16 ------(2')
Now rearranging the terms, we get:
h^2 - 2h + k^2 - 4k - 11 = 0 ------(3)
m^2 - 2m + n^2 - 4n - 11 = 0 ------(4)
Since the two circles are tangent to the same graph, they must have the same equation. Therefore, equations (3) and (4) are identical:
h^2 - 2h + k^2 - 4k - 11 = m^2 - 2m + n^2 - 4n - 11
Rearranging terms, we get:
h^2 - m^2 - 2h + 2m + k^2 - n^2 - 4k + 4n = 0
(h^2 - m^2) + (2m - 2h) + (k^2 - n^2) + (4n - 4k) = 0
(h + m) (h - m) + 2(m - h) + (k + n) (k - n) + 4(n - k) = 0
(h + m - 2) (h - m) + (k + n - 4) (k - n) = 0
Now, we can rewrite this equation as:
(h + m - 2) (h - m) = (4 - k - n) (k - n)
Since the left side and right side of the equation are equal, their factors must be equal as well:
h + m - 2 = 4 - k - n -----(5)
h - m = k - n ------(6)
Now we have a system of equations (5) and (6) that we can solve to find the values of h, k, m, and n.
From equation (6), we can express h in terms of m, k, and n:
h = m + (k - n) ----- (7)
Substituting equation (7) into equation (5), we have:
m + (k - n) + m - 2 = 4 - k - n
2m + 2k - 2n - 6 = 0
Dividing through by 2, we get:
m + k - n - 3 = 0 ----(8)
Now we have a relationship between m, k, and n as given by equation (8). Let's solve for n:
n = m + k - 3
Substituting this value for n into equation (7), we can express h in terms of m and k:
h = m + (k - (m + k - 3))
Simplifying this equation, we get:
h = m + (3 - m) = 3
So, the x-coordinate of the centers of the circles is 3.
Using equation (6), we have h - m = k - n. Substituting h = 3 and n = m + k - 3, we get:
3 - m = k - (m + k - 3)
Simplifying this equation, we get:
3 - m = -2m + 3
Rearranging terms, we have:
m = 0
So, the y-coordinate of the centers of the circles is 0.
Therefore, the centers of the two circles are (3, 0) and (3, 0). Since they have the same centers, the equations of the two circles will be the same.
Using the center and the radius, we can write their equations as:
(x - 3)^2 + (y - 0)^2 = 4^2
(x - 3)^2 + y^2 = 16
Therefore, the equations of the two circles are:
(x - 3)^2 + y^2 = 16.
To find the equations of the two circles, we need to determine their centers. Given that the circles are tangent to the graph y^2 = 4x at the point (1,2), we can set up a system of equations using the center-point form of a circle.
Let (h, k) be the coordinates of the centers of the two circles. The equations of the circles can then be written as:
Circle 1: (x - h)^2 + (y - k)^2 = 4^2
Circle 2: (x - h)^2 + (y - k)^2 = 4^2
Since both circles have the same radius, the equations for both circles will be the same except for the centers (h, k).
Now, let's find the values of h and k.
Since the point (1, 2) lies on the graph y^2 = 4x and is also tangent to the circles, it should satisfy the equation of the circle:
(1 - h)^2 + (2 - k)^2 = 4^2
Expanding this equation gives us:
1 - 2h + h^2 + 4 - 4k + k^2 = 16
Rearranging the terms and simplifying the equation further, we have:
h^2 - 2h + k^2 - 4k - 11 = 0
This equation represents a circle with center (h, k) and radius √11.
Now, to find the equations of the circle, we need to find the values of h and k that satisfy both the equation of the circle and the graph y^2 = 4x.
By substituting y^2 = 4x in the equation of the circle, we get:
h^2 - 2h + (4k - 4√x) - 11 = 0
Substituting x = 1 and y = 2, we have:
h^2 - 2h + (4k - 4√1) - 11 = 0
h^2 - 2h + 4k - 4 - 11 = 0
h^2 - 2h + 4k - 15 = 0
Now, we have a system of equations involving two unknowns, h and k, formed by the equations:
h^2 - 2h + k^2 - 4k - 11 = 0
h^2 - 2h + 4k - 15 = 0
To solve this system, we can use substitution or elimination. Solving the system, we find:
h = 1, k = 3
Now that we have the values of h and k, we can substitute them back into the equation of the circle to find the equations of the two circles:
Circle 1: (x - 1)^2 + (y - 3)^2 = 4^2
Circle 2: (x - 1)^2 + (y - 3)^2 = 4^2
Therefore, the equations of the two circles are:
Circle 1: (x - 1)^2 + (y - 3)^2 = 16
Circle 2: (x - 1)^2 + (y - 3)^2 = 16