Question 739955: Find the equation of the tangent at the point (0,3) on a circle of x sqr + y sqr = 1.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find the equation of the tangent at the point (0,3) on a circle of x sqr + y sqr = 1.
-------------
Use ^ (Shift 6) for exponents.
x^2 + y^2 = 1
(0,3) is not on the circle.
If you mean thru (0,3) and tangent to the circle:
-----
(0,3) is P and the tangent point is T
O is the origin
OTP is a right triangle with T = 90 degs
side TP = sqrt(3^2 - 1^2) = sqrt(8)
----
The tangent point is on the given circle and on the circle centered at (0,3) with radius sqrt(8) --> x^2 + (y-3)^2 = 8
---------
Find the intersections of the 2 circles (2 of them)
x^2 + (y-3)^2 = 8
x^2 + y^2 = 1
----------------------- Subtract
(y-3)^2 - y^2 = 7
y^2 - 6y + 9 - y^2 = 7
-6y = -2
y = 1/3
-------
x = -sqrt(8)/3 --> (-sqrt(8)/3,1/3)
x = +sqrt(8)/3 --> (+sqrt(8)/3,1/3)
------------------
Find the eqn of the line thru the pairs of points
---
|x y 1|
|0 3 1| = 0
|sq 1/3 1|
sq is the point (sqrt(8)/3,1/3)
------------
In Quadrant 1:
x*(3 - 1/3) - y*(0 - sqrt(8)) + (0 - sqrt(8)) = 0
8x/3 + sqrt(8)y = sqrt(8)
--------------------------
You can find the other line.
|
|
|
