# Math Help - circle geometry, exam tmw!

1. ## circle geometry, exam tmw!

A tangent is drawn to the circle x2 + y2 - 18x + 4y + 35 = 0 at he point (14,3).

i) finf the equation to the tangent.
I already did, its y + x - 17 = 0.

ii) find the equations of the two chords, each of length 6(squareroot of 2), which are parallel to the tangent.

2. Originally Posted by milan
A tangent is drawn to the circle x2 + y2 - 18x + 4y + 35 = 0 at he point (14,3).
Complete the square to find the center of the circle.

Originally Posted by milan
i) finf the equation to the tangent.
Find the slope of the line through the center and the point of tangency. Note that the tangent will be perpendicular to radius line of the circle at the point of tangency, so the perpendicular slope is the slope of the tangent.

Then plug the perpendicular slope and the point of tangency into one of the formulas for equations of straight lines to find the tangent line's equation. (I get the same equation that you did.)

Originally Posted by milan
ii) find the equations of the two chords, each of length 6(squareroot of 2), which are parallel to the tangent.
You have the tangent-line's slope, so you know the slope of the two chords. From the circle equation, you know that the radius is 5*sqrt[2].

There's probably a "nicer" way to do this, but one method might be as follows:

The line from the center to the midpoint of one of the chords will have a slope of m = 1. Find the equation, through the center, of the line with this slope.

Draw the circle with its two chords. (The particular scale isn't important. The drawing is just a helpful way of keeping track of stuff.) Draw the line from the center to the midpoint of one of the chords. Also draw the radius lines from the center to either end of this chord. You should now have two identical right triangles.

Label the side of a triangle along the chord as 3sqrt[2] and the side corresponding to a radius line as 5sqrt[2]. Use the Pythagorean Theorem to find the length of the remaining side. This gives the distance of the chord from the center of the circle.

Use the Distance Formula to find the points on this line that are this distance from the center of the circle.

Then find the equations of the lines containing the chords, using the points you just found with the slope you already know.

If you get stuck, please reply showing how far you have gotten. Thank you!

3. Hello, milan!

A tangent is drawn to the circle: . $x^2 + y^2 - 18x + 4y + 35 \:= \:0$ at the point $P(14,3).$

1) Find the equation to the tangent.
I already did . . . it is: . $y + x - 17 \:= \:0$

2) Find the equations of the two chords, each of length $6\sqrt{2}$, which are parallel to the tangent.

We complete the square and get: . $(x-9)^2 + (y+2)^2 \:=\:50$

Make sketch . . .
Code:
         Figure 1
|                 o
|                   o
|               * * * o
|           *           *  P(14,3)
|         *               o
|        *              *  *o
- - + - - - - - - - - - - * - - - o
|       *           *       *   o
|       *       C *         *
|       *       (9,-2)      *
|
|        *                 *
|         *               *
|           *           *
|               * * *
|
The center of the circle is $C(9,\text{-}2)$
Its radius is $PC = 5\sqrt{2}$.

The equation of the tangent is: . $y \:=\:-x+17$
And its slope is -1.

A new diagram to examine one of the chords.
I rotated the diagram so the chord is horizontal.
Code:
      Figure 2

* * * _
Q  *     A 3√2 *  P
* - - - + - - - *
*  *     |     * _*
*   |   * 5√2
*       * | *       *
* - - - - * - - - - *
C

The radius is: $CP = 5\sqrt{2}$
The chord is: $PQ = 6\sqrt{2} \quad\Rightarrow\quad AP = 3\sqrt{2}$

We see that $\Delta PAC$ is a 3-4-5 right triangle. .Hence: . $AC = 4\sqrt{2}$
. . That is, the chord is $\tfrac{4}{5}\text{(radius)}$ from the center.

Look at Figure 1.

Point $A$ is $\tfrac{4}{5}$ of the way from $C(9,\text{-}2)$ to $P(14,3)$.
. . Hence, point $A$ is at $(13,2)$.
The other chord cross the diameter at $B(5,\text{-}6)$.

We know all about the two chords!

. . One contains $A(13,2)$ and has slope $-1$.
. . The other has $B(5,\text{-}6)$ and has slope $-1$.

Go for it!