Find the equations of the straight lines passing through the origin that are tangent to the circle of the center (2, 1) and radius 2

Question

Find the equations of the straight lines passing through the origin that are tangent to the circle of the center (2, 1) and radius 2

Answer 1

As Raymond points out, there are many ways to approach this problem, and I enjoyed his mention of graphing and seeing one solution.

The way I thought of was that the points of tangency create 2 right triangles where the hypotenuse is from (0, 0) to (2, 1), which has length √5 and one leg of the right triangle from the point of tangency to the center of the circle is of length 2.

Thus, the distance from the origin to the point of tangencies will be 1:

Thus, if the point of tangency is (x,y),

x^2 + y^2 = 1 (from the origin to the tangent)

(x-2)^2 + (y-1)^2 = 4 (from the tangency to the center

x^2 - 4x + 4 + y^2 - 2y + 1 = 4

Then, substituting 1 for x^2 + y^2,

-4x - 2y = -2, or

2x + y = 1

y = 1 - 2x

substituting into x^2 + y^2 = 1,

x^2 + (1 - 2x)^2 = 1

x^2 + 1 - 4x + 4x^2 = 1

5x^2 - 4x = 0

x(5x - 4) = 0

x = 0, 4/5

y = 1 - 2x

y = 1, -3/5

The points of tangency are (0, 1) and (4/5, -3/5)

The equations of the lines from the origin (0, 0) through these points is

x = 0 and y = -3/4x, respectively.