How do I find the point P on a circle with a slope of 2 from (0,0)?

[mathdad]

Find a point P on the circle x^2 + y^2 = 20 such that the slope of the radius drawn from (0, 0) to P is 2. (You will get two answers.)

There are no sample questions in the book for me to follow.

What are the steps needed for me to solve this problem?

Why two answers?

 

[MarkFL]

Consider the circle:

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

and the line:

\(\displaystyle y=m(x-h)+k\)

This line passes through the center of the circle, and so will intersect the circle at two points (and the slope of the two resulting radii will be $m$). You have two equations in two unknowns...what do you find?

 

[mathdad]

Consider the circle:

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

and the line:

\(\displaystyle y=m(x-h)+k\)

This line passes through the center of the circle, and so will intersect the circle at two points (and the slope of the two resulting radii will be $m$). You have two equations in two unknowns...what do you find?

Are you saying that I must plug the value of y as given in the second equation into (x - h)^2 + (y - k)^2 = r^2?

 

[MarkFL]

Are you saying that I must plug the value of y as given in the second equation into (x - h)^2 + (y - k)^2 = r^2?

I would observe that the equation of the line may be arranged as:

\(\displaystyle y-k=m(x-h)\)

Now, plug into the equation of the circle for $y-k$...to get:

\(\displaystyle (x-h)^2+m^2(x-h)^2=r^2\)

Now, you can solve this for $x$, and from this value of $x$ determine the value of $y$...what do you get?

 

[mathdad]

I would observe that the equation of the line may be arranged as:

\(\displaystyle y-k=m(x-h)\)

Now, plug into the equation of the circle for $y-k$...to get:

\(\displaystyle (x-h)^2+m^2(x-h)^2=r^2\)

Now, you can solve this for $x$, and from this value of $x$ determine the value of $y$...what do you get?

I am on the train now. I will work on this later or tomorrow.

 

[mathdad]

I am on the train now. I will work on this later or tomorrow.

There is no (y - k) in your equation of the circle not centered at the origin? Is this a typo?

 

[MarkFL]

There is no (y - k) in your equation of the circle not centered at the origin? Is this a typo?

No, I made a substitution...

 

[mathdad]

I will work on this question later. On the bus going home now. I need to solve for x. After doing so, what is the next step? What does solving for x yield in this case?

 

[MarkFL]

I will work on this question later. On the bus going home now. I need to solve for x. After doing so, what is the next step? What does solving for x yield in this case?

What I would is with:

\(\displaystyle (x-h)^2+m^2(x-h)^2=r^2\)

Solve for $x-h$ and plug that into:

\(\displaystyle y=m(x-h)+k\)

to get the $y$-coordinates. Then solve the equation you have for $x-h$ for $x$, and you'll have the $x$-coordinates...make sure you pair the corresponding coordinates correctly.

Now you'll have formulas for the coordinates...you just need to identify $h,k,r,m$ in the specific given problem.

 

[mathdad]

(x - h)^2 + m^2(x - h)^2 = r^2

(x - h)^2[1 + m^2] = r^2

(x - h)^2 = r^2/(1 + m^2)

Taking the square root on both sides, we get

x - h = r/[sqrt(1 + m^2)]

y = m(x - h) + k

y = m(r/[sqrt(1 + m^2)]) + k

 

[MarkFL]

(x - h)^2 + m^2(x - h)^2 = r^2

(x - h)^2[1 + m^2] = r^2

(x - h)^2 = r^2/(1 + m^2)

Taking the square root on both sides, we get

x - h = r/[sqrt(1 + m^2)]

y = m(x - h) + k

y = m(r/[sqrt(1 + m^2)]) + k

 

[mathdad]

Is the picture indicating confusion at my end?

 

[MarkFL]

Is the picture indicating confusion at my end?

No, it is indicating difficulty on my end in reading plain text math. :)

 

[mathdad]

Can you solve this problem for me? Also, why are my pictures too big for this site but not at the MHF? You are also a member of the MHF, right?

 

[MarkFL]

Can you solve this problem for me?

THis problem will provide both a good opportunity to develop your math skills, and to develop your $\LaTeX$ skills. If you're like me, you will learn a great deal more by doing rather than watching.

Also, why are my pictures too big for this site but not at the MHF? You are also a member of the MHF, right?

We have reasonable image size restrictions in place here. I don't recall anyone before having any issue with it. If you find an image is too large, then use an image editor to either shrink the dimensions, or reduce the image quality so that the size of the file is reduced.

Using images in place of $\LaTeX$, except in rare cases where a diagram is included, isn't what I would consider a viable alternative. You will eventually run into a limitation on the total amount of images you are allowed to upload, and it wouldn't allow people to simply edit a quoting of your post to correct any math.

I have been involved with math help for over 7 years now, and I have never seen anyone fight using $\LaTeX$ so much. It is something you really should take the time to do. :)

 

[mathdad]

I will try to upload a picture reply. If not, then I will try LaTex. Honestly, LaTex delays my typing work.