In polar coordinates, the parametric equations x=6+cos(f) and y=sin(f) represent a circle C1. In Cartesian coordinates, there is a circle C2 that is externally tangent to C1, tangent to the y-axis, and centered at (10,sqrt2). What is the value of a?
I assume a is the radius of C2.
since cos^2+sin^2 = 1,
C1 is (x-6)^2 + y^2 = 1
The distance between the centers of C1 and C2 is thus
d^2 = (10-6)^2 + (√2)^2 = 18
so, a+1 = √18, and a = √18 - 1
I made a small typing mistake. The center is (10,sqrta). so, what is a?
If C2 has center at (10,y) and touches the y-axis, its radius is 10.
Then distance between the circle centers is now
√(a+16) = 10+1 = 11
a+16 = 121
a = 105
To find the value of a, we need to determine the radius of the circle C2. Let's break down the problem step by step:
1. Start by finding the point of tangency between the circles C1 and C2. Since C2 is tangent to the y-axis, the x-coordinate of this point will be equal to the radius of C2. Let's call this point (a, b).
2. Now, let's substitute (a, b) into the equation of C1 to find the corresponding polar coordinates. Using x = 6 + cos(f) and y = sin(f), we substitute the x-coordinate (a) into the equation x = 6 + cos(f) and the y-coordinate (b) into the equation y = sin(f).
Substituting a into x, we have:
a = 6 + cos(f)
Substituting b into y, we have:
b = sin(f)
3. To convert these polar coordinates to Cartesian coordinates, we can use the following relations:
x = r * cos(f)
y = r * sin(f)
Comparing these equations with the equations from step 2, we can determine that r = 6.
4. Now, since we know the center of C2 is (10, sqrt(2)), we can calculate the distance from the center of C2 to the point of tangency on the x-axis. This distance is the radius of C2.
Using the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Substituting (10, sqrt(2)) as (x1, y1) and (a, 0) as (x2, y2), the formula becomes:
Radius = sqrt((a - 10)^2 + (0 - sqrt(2))^2)
5. Now, equate the radius of C2 (from step 4) with the calculated value from step 3:
sqrt((a - 10)^2 + (0 - sqrt(2))^2) = 6
6. Square both sides of the equation to remove the square root:
(a - 10)^2 + (0 - sqrt(2))^2 = 36
7. Simplify the equation:
(a - 10)^2 + 2 = 36
8. Move 2 to the other side:
(a - 10)^2 = 34
9. Take the square root of both sides to isolate (a - 10):
a - 10 = ±sqrt(34)
10. Finally, add 10 to both sides:
a = 10 ± sqrt(34)
So, the value of a is 10 ± sqrt(34).