# Thread: Intersect of Line and Circle

1. ## Intersect of Line and Circle

Samatha is running near the Circular Park, the shape of a perfect circle. It has a radius of 8 cm. She begins from a point 10 cm west and 3 cm south of the center of the park. She heads toward a point 20 cm east and 4 cm north of the center of the park. Though, when she reaches a point due east of the center of the forest, she turns and runs due south until she exits the park.

Samatha runs at a constant 5 cm per hour. How much time did she spend in the forest?

So the things I bolded were the main given facts and numbers
So I tried drawing a coordinate plane.
1) I made the center of the forest (0,0)
2) She begins from point 10 cm west and 3 cm south, which means she began at (10, -3)
3) She heads toward a point 20 cm east and 4 cm north of the center of the forest, so it's at (20,4), but she's heading there
4) I don't know if I'm correct here, she moves east now to the center of the forest then runs south?
Then (5) would be another step,
Would I use the standard equation of the circle? x2+y2=r2, r being the radius

If someone could help me by providing the steps or explaination, that would be nice
Thanks!

2. ## Re: Intersect of Line and Circle

Originally Posted by Chaim
Samatha is running near the Circular Park, the shape of a perfect circle. It has a radius of 8 km. She begins from a point 10 cm west and 3 cm south of the center of the park. She heads toward a point 20 cm east and 4 km north of the center of the park. Though, when she reaches a point due east of the center of the forest, she turns and runs due south until she exits the park.

Samatha runs at a constant 5 cm per hour. How much time did she spend in the forest?

So the things I bolded were the main given facts and numbers
So I tried drawing a coordinate plane.
1) I made the center of the forest (0,0)
2) She begins from point 10 cm west and 3 km south, which means she began at (10, -3)
3) She heads toward a point 20 cm east and 4 km north of the center of the forest, so it's at (20,4), but she's heading there
4) I don't know if I'm correct here, she moves east now to the center of the forest then runs south?
Then (5) would be another step,
Would I use the standard equation of the circle? x2+y2=r2, r being the radius

If someone could help me by providing the steps or explaination, that would be nice
Thanks!
you do realize that 5 cm/hr is centimeters per hour ? a snail moves faster.

sure you don't mean km (kilometers) instead of cm (centimeters) everywhere in this problem?

3. ## Re: Intersect of Line and Circle

check distances and rate of running. cm per hour ????

4. ## Re: Intersect of Line and Circle

Ah, it's just the units
But yeah, I accidently messed up, it's all suppose to be the same units, so use centimeters xD

5. ## Re: Intersect of Line and Circle

she starts at (-10,-3) , not (10,3)

she heads toward the point (20,4)

start by finding the equation of the line between (-10,-3) and (20,4)

once you find that linear equation, solve for either x or y (whichever is easier) and substitute the result into the circle equation, then solve for the single variable ... understand that the line first intersects the circle in quadrant III (x and y both negative).

you also need to determine where the line crosses the x-axis (x-intercept), because that is where she turns to head due south.

attached is a sketch of the problem ... the red line segments mark the path she travels.

... one final tip, use kilometers for distance.

6. ## Re: Intersect of Line and Circle

Originally Posted by skeeter
she starts at (-10,-3) , not (10,3)

she heads toward the point (20,4)

start by finding the equation of the line between (-10,-3) and (20,4)

once you find that linear equation, solve for either x or y (whichever is easier) and substitute the result into the circle equation, then solve for the single variable ... understand that the line first intersects the circle in quadrant III (x and y both negative).

you also need to determine where the line crosses the x-axis (x-intercept), because that is where she turns to head due south.

attached is a sketch of the problem ... the red line segments mark the path she travels.

... one final tip, use kilometers for distance.
Lol okay I'll use kilometers
And oopsies, yeah, I keep making typos xD, so yeah it's (-10,-3)
Ooo I see! So basically I use y=mx+b right?
Using (-10, -3) and (20,4) since she's making a straight line there?
So (4-(-3))/(20-(-10))=(7/30)
Then y=(7/30)x+b
So using (-10, -3) to find b
-3=(7/30)(10)+b
-3=7/3+b
-9/3=7/3+b
-16/3=b
So y=(7/30)x-(16/3)
Then to find y intercept -> y=(7/30)(0)-(16/3) -> y=-16/3
Then to find x-intercept -> 0=(7/30)x-(16/3)-> 16/3=(7/30)x -> x = 112/90
Just making sure, am I doing this right so far?

7. ## Re: Intersect of Line and Circle

Hello, Chaim!

I will assume that all the measurements are in kilometers.
I'll outline the Game Plan for this problem.

Samatha is running through Circular Park, in the shape of a perfect circle with a radius of 8 km.
She begins from a point 10 km west and 3 km south of the center of the park.
She heads toward a point 20 km east and 4 km north of the center of the park.
When she reaches a point due east of the center of the park,
. . she turns and runs due south until she exits the park.

Samatha runs at a constant 5 km per hour. .How much time did she spend in the park?

Code:
                        |
* * *                          (20.4)
*     |     *                         o B
*       |       *                 *     :
*        |        *          *           :
|               *               :
*         |   Q     *                     :
- - * - - - - * - - - - + - o - - * - - - - - - - - - - * - -
:         *       * |   |     *
:       P o *       |   |
:     *    *        |   |    *
A o           *       |   |   *
(-10,-3)          *     |   o *
* * *
|   R
The equation of the park is: . $x^2 + y^2 \,=\,64$

Samantha starts at $A(\text{-}10,\text{-}3)$ and runs straight to $B(20,4).$
The equation of line $AB$ is: . $y \:=\:\tfrac{7}{30}x - \tfrac{2}{3}$

She enters the park at point $P.$
We must find the first intersection of line $AB$ and the circle.

She cross the x-axis at point $Q.$
We must find the x-intercept, $x_o$, of line $AB.$

Then she runs directly south and exits the park at point $R.$
We must determine $y$ (in the circle) when $x = x_o$

Then we find her total distance in the park, $PQ + QR$ km,
. . and divide by her speed, 5 km/hr.

This gives us her time spent in the park.

Ah! . . . Skeeter already explained all this . . . *sigh*

8. ## Re: Intersect of Line and Circle

Once you get the equation as shown by Soroban the two running distances are easily solved by PYt theorem .The turning point where y=0 is 60/21

9. ## Re: Intersect of Line and Circle

Originally Posted by Soroban
Hello, Chaim!

I will assume that all the measurements are in kilometers.
I'll outline the Game Plan for this problem.

Code:
                        |
* * *                          (20.4)
*     |     *                         o B
*       |       *                 *     :
*        |        *          *           :
|               *               :
*         |   Q     *                     :
- - * - - - - * - - - - + - o - - * - - - - - - - - - - * - -
:         *       * |   |     *
:       P o *       |   |
:     *    *        |   |    *
A o           *       |   |   *
(-10,-3)          *     |   o *
* * *
|   R
The equation of the park is: . $x^2 + y^2 \,=\,64$

Samantha starts at $A(\text{-}10,\text{-}3)$ and runs straight to $B(20,4).$
The equation of line $AB$ is: . $y \:=\:\tfrac{7}{30}x - \tfrac{2}{3}$

She enters the park at point $P.$
We must find the first intersection of line $AB$ and the circle.

She cross the x-axis at point $Q.$
We must find the x-intercept, $x_o$, of line $AB.$

Then she runs directly south and exits the park at point $R.$
We must determine $y$ (in the circle) when $x = x_o$

Then we find her total distance in the park, $PQ + QR$ km,
. . and divide by her speed, 5 km/hr.

This gives us her time spent in the park.

Ah! . . . Skeeter already explained all this . . . *sigh*
Ah.. I see! Ok thanks!