New Reply

Equation of a Circle with a Center and Tangent Point

 
Share Thread Thread Tools
Jul30-12, 09:23 PM   #1
 

Equation of a Circle with a Center and Tangent Point


What is the equation of the circle with a center point of (10, -14) when the circle is tangent to x=13?

D = √(13-10)^2 + (0-(14))^2
D = √(3)^2 + (14))^2
D = √9+196
D = √205

Radius = √205
(x-10)^2 + (y-(-14))^2 = √205^2
(x-10)^2 + (y+14)^2 = 205

But how am I suppose to graph this?
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> Galaxies fed by funnels of fuel
>> The better to see you with: Scientists build record-setting metamaterial flat lens
>> Google eyes emerging markets networks
Jul30-12, 09:31 PM   #2
 
Before we can help you, you need to show us what you have tried already.
Jul30-12, 09:48 PM   #3
 
Recognitions:
Homework Helper Homework Help
Quote by xxmegxx View Post
What is the equation of the circle with a center point of (10, -14) when the circle is tangent to x=13?

D = √(13-10)^2 + (0-(14))^2
D = √(3)^2 + (14))^2
D = √9+196
D = √205

Radius = √205
(x-10)^2 + (y-(-14))^2 = √205^2
(x-10)^2 + (y+14)^2 = 205

But how am I suppose to graph this?
Your method of calculating the radius (if D is supposed to be the radius) makes no sense.

The problem is actually very simple. You're given that the circle is tangent to x=13, which is a vertical line. You know the centre has an x-coordinate of 10. So what can you say about the radius?
Jul30-12, 09:53 PM   #4
 

Equation of a Circle with a Center and Tangent Point


I meant D to be the distance. I'm not sure how you solve this problem without graph paper.
Jul30-12, 09:54 PM   #5
 
Recognitions:
Homework Helper Homework Help
Quote by xxmegxx View Post
I meant D to be the distance. I'm not sure how you solve this problem without graph paper.
Distance from what to what?

There's no need for graph paper. All you need is a reasonable sketch. Remember the general equation for the circle and what the terms represent.
Jul30-12, 09:58 PM   #6
 
The distance from the center point to the tangent line to find the radius.
Jul30-12, 10:08 PM   #7
 
Mentor
Blog Entries: 9
What is the issue with graphing? it is a circle, you know the center and radius. What else do you want?
Jul30-12, 10:19 PM   #8
 
Recognitions:
Homework Helper Homework Help
Quote by xxmegxx View Post
The distance from the center point to the tangent line to find the radius.
But that distance is NOT the radius! Remember that x=13 is a vertical tangent to the circle. A circle with the radius you calculated would not have that line as a tangent.

Also, what you calculated was the distance between the points (10,-14) and (13,0). This is NOT the same as the (shortest) distance between (10,-14) and the line x = 13. Do you see why?

Just do a sketch. Do you see why the radius is simply 13 - 10 = 3?

(As a final point, there was an error in your working in the first post. The distance should've been [itex][(13 - 10)^2 + (0 - (-14))^2]^\frac{1}{2}[/itex]. Note the sign in the y-term. But the squaring masked your error.)
Jul31-12, 07:00 AM   #9
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff Staff Emeritus
Quote by xxmegxx View Post
The distance from the center point to the tangent line to find the radius.
Quote by Curious3141 View Post
But that distance is NOT the radius!
Just to avoid confusion, what he is writing here is the radius. But you are saying that the distance calculated before, from the center of the circle to the point (13, 0), is not "the distance from the center point to the tangent line".

I suspect that the real difficulty is that xxmeqxx is thinking, incorrectly, that "x= 13" means the point on the x-axis with x-component 13 rather than, as every here is telling him, the line of all points whose x-component is 13, (13, y).



Remember that x=13 is a vertical tangent to the circle. A circle with the radius you calculated would not have that line as a tangent.

Also, what you calculated was the distance between the points (10,-14) and (13,0). This is NOT the same as the (shortest) distance between (10,-14) and the line x = 13. Do you see why?

Just do a sketch. Do you see why the radius is simply 13 - 10 = 3?

(As a final point, there was an error in your working in the first post. The distance should've been [itex][(13 - 10)^2 + (0 - (-14))^2]^\frac{1}{2}[/itex]. Note the sign in the y-term. But the squaring masked your error.)
Jul31-12, 07:13 AM   #10
 
Recognitions:
Homework Helper Homework Help
Quote by HallsofIvy View Post
Just to avoid confusion, what he is writing here is the radius. But you are saying that the distance calculated before, from the center of the circle to the point (13, 0), is not "the distance from the center point to the tangent line".
Yes, what he wrote in words (with reference to distance between centre and tangent line) is the radius. But what he calculated (distance between centre and (13,0) ) is not. That's what I meant.
New Reply
Thread Tools


Similar Threads for: Equation of a Circle with a Center and Tangent Point
Thread Forum Replies
Center of circle from two points and a tangent angle Calculus 1
Find the coordinates of a point on a circle without knowing the center point. General Math 2
Equation of a tangent line to a given circle from an external point - Stuck! Precalculus Mathematics Homework 4
How can I find the equation of tangent lines from a circle to point A? Calculus & Beyond Homework 21
Tangent to y axis with center(-3,4) whats the eqn of circle? Introductory Physics Homework 5