Tangent lines to a circle

[toadz0rs]

Hey everyone, this is my first post so I just wanted to say hello before jumping right in. I skimmed through the sticky and think I'm doing this right, if not just point me in the right direction! Now onto my question

"Draw the unit circle and plot the point P=(3,2). Observe there are TWO lines tangent to the circle passing through the point P."

To help visualize, here is the according diagram as well:

I need to find the points of tangency on the circle and their equations through the point (3,2).

I'm struggling finding the points of tangency (once I do, I can construct the equations on my own). I have approached the problem by substituting the line y=mx+b into the circle equation of x^2 + y^2 = 1 and tried to work with the determinant but don't get the correct answer. Is it simpler/possible to approach this problem using angles? I used the distance equation and pythagorean theorem to solve for the length of all three sides of the triangle, but am unsure of how to proceed from there. Please help.

 

[Deleted member 4993]

Hey everyone, this is my first post so I just wanted to say hello before jumping right in. I skimmed through the sticky and think I'm doing this right, if not just point me in the right direction! Now onto my question

"Draw the unit circle and plot the point P=(3,2). Observe there are TWO lines tangent to the circle passing through the point P."

To help visualize, here is the according diagram as well:

I need to find the points of tangency on the circle and their equations through the point (3,2).

I'm struggling finding the points of tangency (once I do, I can construct the equations on my own). I have approached the problem by substituting the line y=mx+b into the circle equation of x^2 + y^2 = 1 and tried to work with the determinant but don't get the correct answer. Is it simpler/possible to approach this problem using angles? I used the distance equation and pythagorean theorem to solve for the length of all three sides of the triangle, but am unsure of how to proceed from there. Please help.

This is one of those problems where construction (ruler-compass) is easier than analytical method. Your approach is basically correct. Should get you the correct answer - but you must be messing up somewhere in algebra. Try the following approach (basically same as yours) and see if you get the correct answer.

Suppose the point of tangency for L[sub:z1bdc2ka]1[/sub:z1bdc2ka]is (x[sub:z1bdc2ka]1[/sub:z1bdc2ka],y[sub:z1bdc2ka]1[/sub:z1bdc2ka])

Then from equation of circle

$\displaystyle \frac{dy}{dx} \ = \ - \ \frac{x}{y}$

then equation for L[sub:z1bdc2ka]1[/sub:z1bdc2ka] is:

$\displaystyle y - y_1 \ = \ - \ \frac{x_1}{y_1}\cdot (x - x_1)$

This line also passes through (3,2) - then:

$\displaystyle 2 - y_1 \ = \ - \ \frac{x_1}{y_1}\cdot (3 - x_1)$

$\displaystyle 2\cdot y_1 - y_1^2 \ = \ - \ 3\cdot x_1 + x_1^2$

$\displaystyle 2\cdot y_1 - y_1^2 \ + \ 3\cdot x_1 - x_1^2 \ = \ 0$

$\displaystyle 2\cdot y_1 \ + \ 3\cdot x_1 \ - \ 1 \ = \ 0$

$\displaystyle y_1 \ = \ \frac{-3\cdot x_1 \ + \ 1}{2}$........................................................(1)

Now use (1) into $\displaystyle x_1^2 \ + \ y_1^2 = 1$

and you'll get the values of co-ordinates of the tangency points of L[sub:z1bdc2ka]1[/sub:z1bdc2ka] and L[sub:z1bdc2ka]2[/sub:z1bdc2ka]

 

[lookagain]

$\displaystyle \ > \ > \frac{dy}{dx} \ = \ - \ \frac{x}{y} \ \ < \ <$

then equation for L[sub:2eb4trou]1[/sub:2eb4trou] is:

$\displaystyle y - y_1 \ = \ - \ \frac{x_1}{y_1}\cdot (x - x_1)$

This line also passes through (3,2) - then:

$\displaystyle 2 - y_1 \ = \ - \ \frac{x_1}{y_1}\cdot (3 - x_1)$

$\displaystyle 2\cdot y_1 - y_1^2 \ = \ - \ 3\cdot x_1 + x_1^2$

$\displaystyle 2\cdot y_1 - y_1^2 \ + \ 3\cdot x_1 - x_1^2 \ = \ 0$

$\displaystyle 2\cdot y_1 \ + \ 3\cdot x_1 \ - \ 1 \ = \ 0$

$\displaystyle y_1 \ = \ \frac{-3\cdot x_1 \ + \ 1}{2}$........................................................(1)

Now use (1) into $\displaystyle x_1^2 \ + \ y_1^2 = 1$

and you'll get the values of co-ordinates of the tangency points of L[sub:2eb4trou]1[/sub:2eb4trou] and L[sub:2eb4trou]2[/sub:2eb4trou]

Subhotosh Khan,

you can't use this method because it uses calculus, as this is the "Geometry and Trig" board.

From a theorem (or lemma) in geometry, the tangent lines are equal in length, the radii
(to the points of tangency) are equal in length (by definition being radii of the same circle),
and they are perpendicular to the lines of tangency.
The distance from the center of the circle, (0,0), to (3,2) is the square root of 13 using the
Pythagorean Theorem.

Then, based on the Pythagorean Theorem, we have:

$\displaystyle (x - 3)^2 + (y - 2)^2 = 13$

And from the equation of the unit circle, we have:

$\displaystyle x^2 + y^2 = 1$


Solving these two equations simultaneously will give the coordinates
of the two points of tangency.