# Thread: Equation of two lines tangent to a circle

1. ## Equation of two lines tangent to a circle

The question:

There are two tangent lines from the point (0,1) to the circle x^2 + (y+1)^2=1 [centered at (0,-1) with a radius of one]. Find equations of these lines by using the fact that each tangent line intersects the circle in exactly one point.

There is also a graph but it shows pretty much what the question says. I have no idea how I can find the slopes of these tangent lines

2. If we solve the circle equation for y, we get $\displaystyle y=\sqrt{1-x^{2}}-1$

We can use $\displaystyle y-y_{1}=m(x-x_{1})$, plug in all of our info wrt x and solve for x.

Where m is the derivative of y, $\displaystyle y_{1}=1, \;\ x_{1}=0$

$\displaystyle \underbrace{\sqrt{1-x^{2}}-1}_{\text{y}}-1=\frac{-x}{\sqrt{1-x^{2}}}(x-0)$

Now, solve for x. There will be the two values of x.

This will be the two x values where the lines are tangent to the circle.

Then, you will have all the info needed to find the line equations.

3. Ah thank you very much!
But I'm wondering if there is a way to find the derivative without using calculus since this problem was in the review section of the calc book?

4. Originally Posted by atruenut
Ah thank you very much!
But I'm wondering if there is a way to find the derivative without using calculus since this problem was in the review section of the calc book?
From the equation of the circle you know that the center of the circle is: C(0, -1).

Let T(x, y) denote the tangent point on the circle then the slope of $\displaystyle \overline{TC}$ is:

$\displaystyle m_{TC}=\dfrac{y-x_C}{x-x_C}=\dfrac{y-(-1)}{x-0}$

Since T is placed on the circle line the coordinates of T are $\displaystyle T(x, \sqrt{1-x^2} - 1)$

Therefore

$\displaystyle m_{TC}=\dfrac{\sqrt{1-x^2} - 1-(-1)}{x}=\dfrac{\sqrt{1-x^2}} x$

The line TC is perpendicular to the tangent because TC = r. Therefore the slope of the tangent must be the negative reciprocal of the slope of TC:

$\displaystyle m_t=-\dfrac x{\sqrt{1-x^2}}$

5. ## Analytic approach

Originally Posted by atruenut
The question:

There are two tangent lines from the point (0,1) to the circle x^2 + (y+1)^2=1 [centered at (0,-1) with a radius of one]. Find equations of these lines by using the fact that each tangent line intersects the circle in exactly one point.

There is also a graph but it shows pretty much what the question says. I have no idea how I can find the slopes of these tangent lines
Let $\displaystyle C:~x^{2}+(y+1)^{2}=1$.
Clearly, the center of the circle is $\displaystyle M=(0,-1)$ and the radius is $\displaystyle r=1$.
Let $\displaystyle Q:=(q_{1},q_{2})$ be an arbitrary point on $\displaystyle C$.
Then, it is easy to see that the tangent line of the circle at $\displaystyle Q$ is given by
$\displaystyle \langle\underset{\overrightarrow{QX}}{\underbrace{ (x-q_{1},y-q_{2})}},\underset{\overrightarrow{MQ}}{\underbrac e{(q_{1},q_{2}+1)}}\rangle=0$
or simply
$\displaystyle q_{1}(x-q_{1})+(y-q_{2})(q_{2}+1)=0$.......................(*)
We want this line pass through the point $\displaystyle (0,1)$.
Then, we have $\displaystyle 0=q_{1}^{2}+(q_{2}-1)(q_{2}+1)=q_{1}^{2}+q_{2}^{2}-1$, using the fact that $\displaystyle Q\in C$, we have $\displaystyle q_{1}^{2}+q_{2}^{2}=-2q_{2}$, therefore, we obtain $\displaystyle q_{2}=-1/2$.
Using this in the circle's equation, we have $\displaystyle q_{1}=\pm\sqrt{3}/2$.
Using $\displaystyle Q=(-1/2,\pm\sqrt{3}/2)$ in (*), we get the desired tangent line equations.

6. Hello, atruenut!

There are two tangent lines from the point (0,1) to the circle $\displaystyle x^2 + (y+1)^2\:=\:1$

Find equations of these tangents.
Code:
                |
P * (0,1)
/|\
.         / | \
/  |  \
/   |   \
/    |    \
- - - - / - * * * - \ - - - - - -
/*     |     *\
o       |       o A
*  *     |     *  *
*   |   *
*       * | *       *
*       C *(0,-1)   *
*         |         *
|
*        |        *
*       |       *
*     |     *
* * *
|

The center of the circle is: $\displaystyle C(0,-1)$
Its radius is: $\displaystyle CA = 1$

A radius is perpendicular to a tangent at the point of tangency.
. . Hence: .$\displaystyle \angle CAP \:=\:90^o$

In right triangle $\displaystyle CAP$ we have: .$\displaystyle CA = 1,\;CP = 2$
. . Hence, $\displaystyle \Delta CAP$ is a 30-60 right triangle . . . $\displaystyle \angle APC = 30^o$

You should be able to determine the slope of $\displaystyle PA$ . . . right?

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# what is the equation of a circle with two tangent line

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