# GR11 Math: finding endpoints of chord within a circle.

• May 6th 2009, 06:43 PM
nathan02079
GR11 Math: finding endpoints of chord within a circle.
Given...

1. Point P is on the line $\displaystyle 3x+y=26$ and is 10 units from the origin. Determine the coordinates of P.

I've gotten this far...

So:
$\displaystyle 3x+y=26$
$\displaystyle x^2+y^2=\sqrt{10}$.

Then $\displaystyle y=26-3x$.

Plugging in y...

$\displaystyle x^2+(26-3x)^2=\sqrt{10}$

$\displaystyle 10x^2-156x+676=\sqrt{10}$

This is where I'm stuck since I cannot factor it. When the quadratic formula is applied, the number inside the brackets is a negative number.

2.
Given...

From a lighthouse, the range of visibility on a clear day is 40km. On a coordinate system, where O(0,0) represents the lighthouse, a ship is travelling on a course represented by $\displaystyle y=2x+80$. Between which two points on the course can the ship be seen from the lighthouse?

I got this far...

$\displaystyle y=2x+80$
$\displaystyle x^2+y^2=\sqrt{40}$

Plugging it in...

$\displaystyle x^2+(2x+80)^2=\sqrt{40}$

$\displaystyle 5x^2+320x+6400=\sqrt{40}$

Again, I am stuck with the same problem. I get a negative number when the quadratic formula is applied.

What am I doing wrong here? Thank you so much!

N.
• May 6th 2009, 07:45 PM
pickslides
Quote:

Originally Posted by nathan02079
Given...

1. Point P is on the line $\displaystyle 3x-y=26$ and is 10 units from the origin. Determine the coordinates of P.

I've gotten this far...

So:
$\displaystyle 3x+y=26$

(Itwasntme)

Are these signs supposed to be different?
• May 6th 2009, 07:46 PM
Soroban
Hello, nathan02079!

In your "distance formula", you're square-rooting when you should square.

Quote:

2. From a lighthouse, the range of visibility on a clear day is 40 km.
On a coordinate system, where O(0,0) represents the lighthouse,
a ship is travelling on a course represented by: $\displaystyle y\:=\:2x+80$.
Between which two points on the course can the ship be seen from the lighthouse?

Code:

                        |                         |    *                         | *                   Q  * |                   o    |               *    *    |       P  *        *  |       o              *  |   *        *        * |                   *    *|   - - - - - - - - - - - * - - - -                         |O

$\displaystyle P$ and $\displaystyle Q$ have coordinates $\displaystyle (x,y)$ such that: .$\displaystyle OP = OQ = 10.$

We want: .$\displaystyle x^2 + y^2 \:=\:10^2$

So we have: .$\displaystyle x^2 + (2x+80)^2 \:=\:10^2 \quad\Rightarrow\quad 5x^2 + 320x + 4800 \:=\:0$

Factor: .$\displaystyle 5(x+40)(x+24) \:=\:0$

Then: . $\displaystyle x \:=\:\text{-}40,\:\text{-}24 \quad\Rightarrow\quad y \:=\:0,\:32$

The ship can be seen from $\displaystyle (\text{-}40,\:0)$ to $\displaystyle (\text{-}24,\:32)$

• May 6th 2009, 08:44 PM
nathan02079
Quote:

Originally Posted by pickslides
(Itwasntme)

Are these signs supposed to be different?

Whoops, typo there. :/

Quote:

Originally Posted by Soroban
Hello, nathan02079!

In your "distance formula", you're square-rooting when you should square.

Code:

                        |                         |    *                         | *                   Q  * |                   o    |               *    *    |       P  *        *  |       o              *  |   *        *        * |                   *    *|   - - - - - - - - - - - * - - - -                         |O

$\displaystyle P$ and $\displaystyle Q$ have coordinates $\displaystyle (x,y)$ such that: .$\displaystyle OP = OQ = 10.$

We want: .$\displaystyle x^2 + y^2 \:=\:10^2$

So we have: .$\displaystyle x^2 + (2x+80)^2 \:=\:10^2 \quad\Rightarrow\quad 5x^2 + 320x + 4800 \:=\:0$

Factor: .$\displaystyle 5(x+40)(x+24) \:=\:0$

Then: . $\displaystyle x \:=\:\text{-}40,\:\text{-}24 \quad\Rightarrow\quad y \:=\:0,\:32$

The ship can be seen from $\displaystyle (\text{-}40,\:0)$ to $\displaystyle (\text{-}24,\:32)$

Wow, duh! Thanks for stating the obvious! Darn, I never pay attention.

Thank you!

N.

SOLUTION #1:

$\displaystyle 10x^2-156x+576=0$
$\displaystyle x=9.6, 6$

Points of intersection: (9.6, -2.8), (6,8)

SOLUTION #2:

$\displaystyle 5x^2+320x+4800=0$
$\displaystyle x=-24, -40$

Points of intersection: (-24, 32), (-40,0)

Thanks again!