Unit circle and tangent line

[EV33]

Homework Statement

We are given the unit circle and the point (5,2). There are two lines that are tangent to the unit circle and they both intersect at the point (5,2). What are the points where these lines are tangent with the unit circle.

Homework Equations

Tangent line of a circle at the point x,y will have a slope of -x/y
y=ax+b is a linear line

The Attempt at a Solution

y=ax+b

Let's call the tangent point (u,v)
Thus a=-(u/v)

y=-(u/v)x+b
2=-(u/v)5+b
b=2+(u/v)5
y=-(u/v)x+2+5(u/v)
=(u/v)(5-x)+2

Then from here I am stuck because when I go to solve for u/v by plugging in the point I lose my u/v. I feel like I am close. Could I get a hint please.

Thank You.

 

[Mark44]

Homework Statement

We are given the unit circle and the point (5,2). There are two lines that are tangent to the unit circle and they both intersect at the point (5,2). What are the points where these lines are tangent with the unit circle.

Homework Equations

Tangent line of a circle at the point x,y will have a slope of -x/y
y=ax+b is a linear line

The Attempt at a Solution

y=ax+b

Let's call the tangent point (u,v)
Thus a=-(u/v)

y=-(u/v)x+b
2=-(u/v)5+b
b=2+(u/v)5
y=-(u/v)x+2+5(u/v)
=(u/v)(5-x)+2

Then from here I am stuck because when I go to solve for u/v by plugging in the point I lose my u/v. I feel like I am close. Could I get a hint please.

Thank You.

As stated in the problem, there are two tangent points on the circle, and you have identified them as (u, v).

What you didn't take into account was the equation of the circle, which is u² + v² = 1, still using your variables. By solving for, say, v in terms of u, you can eliminate one of your variables.

Any point (u, v) on the circle can be written as (u, +/-sqrt(1 - u²)).

Let's call the tangent point in the 2nd quadrant (-u, +sqrt(1 - u²), where u > 0. The other tangent point in the 3rd quadrant is (u, -sqrt(1 - u²)), where again u > 0.

Take each of the tangent points and calculate the slope of the segment between that point and (5, 2), and set this slope equal to the slope of the tangent on the circle. That should give you an equation in u that you can solve.