Equation of a Circle with a Center and Tangent Point

In summary, the equation of a circle with a center and tangent point can be written as (x - h)^2 + (y - k)^2 = r^2. To find the equation, you need to know the coordinates of the center and one point on the circle and use the formula. There can only be one equation of a circle with a given center and tangent point. To graph a circle, you can plot the center point and use a ruler to draw the radius. The equation is significant in geometry and helps us understand the relationship between the center, tangent point, and radius of a circle.
  • #1
xxmegxx
3
0
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?
 
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  • #2
Before we can help you, you need to show us what you have tried already.
 
  • #3
xxmegxx said:
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?
 
  • #4
I meant D to be the distance. I'm not sure how you solve this problem without graph paper.
 
  • #5
xxmegxx said:
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.
 
  • #6
The distance from the center point to the tangent line to find the radius.
 
  • #7
What is the issue with graphing? it is a circle, you know the center and radius. What else do you want?
 
  • #8
xxmegxx said:
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.)
 
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  • #9
xxmegxx said:
The distance from the center point to the tangent line to find the radius.

Curious3141 said:
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.)
 
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  • #10
HallsofIvy said:
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.
 

FAQ: Equation of a Circle with a Center and Tangent Point

1. What is the equation of a circle with a center and tangent point?

The equation of a circle with a center and tangent point can be written as (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius.

2. How do you find the equation of a circle with a center and tangent point?

To find the equation of a circle with a center and tangent point, you need to know the coordinates of the center and one point on the circle. Then, you can use the formula (x - h)^2 + (y - k)^2 = r^2 and plug in the values to solve for r. Once you have the value of r, you can substitute it into the equation and simplify to get the final equation of the circle.

3. Can you have more than one equation of a circle with a center and tangent point?

No, there can only be one equation of a circle with a given center and tangent point. This is because the equation of a circle is unique and determined by the center and radius of the circle.

4. How do you graph a circle with a center and tangent point?

To graph a circle with a center and tangent point, you can plot the center point (h, k) on a coordinate plane. Then, use a ruler to draw a line from the center point to the tangent point. This line will be the radius of the circle. From the end point of the radius, draw a curved line to complete the circle.

5. What is the significance of the equation of a circle with a center and tangent point?

The equation of a circle with a center and tangent point is significant because it is a fundamental formula in geometry and can be used to solve various problems involving circles. It also helps us understand the relationship between the center, tangent point, and radius of a circle.

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