Equation of a circle from given conditions

In summary: A pleasure. Thanks for your kind compliment - the problem was certainly a little different from 'run of the mill' exercises in analytic geometry.
  • #1
Krushnaraj Pandya
Gold Member
697
73

Homework Statement


Equation of the circle passing through the point (1,2) and (3,4) and touching the line 3x+y-3=0 is?

Homework Equations


x^2+y^2+2gx+2fy+c=0...(1)
(-g,-f)=center of circle
sqrt(g^2+f^2-c)=radius...(2)

The Attempt at a Solution


Putting (1,2) and (3,4) in equation 1 we get 5+2g+4f+c=0; 25+6g+8f+c=0.
Now, line joining the two points will be perpendicular to the line joining center and midpoint of that line (chord perpendicular to radius). Say (h,k) is center, slope joining the two points is 1 so slope of radius through midpoint is -1 (perpendicular lines), midpoint of chord is (2,3); equating -1 to slope of (h,k) and (2,3) gives us k+h=5- but h= -g and k= -f; so -g-f=5 Solving these three equations gives c=40, f= -35/2 and g= 25/2 which is the wrong circle. I know there are other ways to solve this but I want to know why this method is not working in particular- I double checked all the calculations and I can't figure out anything wrong with my logic, Thank you for your help
 
Physics news on Phys.org
  • #2
It is unclear to me what you are trying to do with this ”method”. Why are you creating a line from the centre to the midpoint? You have already used that the two points need to be on the circle and there are an infinite number of circles satisfying this. You cannot squeeze more information out of those two points. You need to use the third requirement.
 
  • Like
Likes Delta2
  • #3
Krushnaraj Pandya said:
so -g-f=5 Solving these three equations gives c=40, f= -35/2 and g= 25/2
Also note that your equation in bold here is not a new equation. You can get it by just using your previous two so there is no new information. Your equation system therefore does not have a unique solution (two equations for three variables) and you need to use the extra information provided.
 
  • #4
Orodruin said:
It is unclear to me what you are trying to do with this ”method”. Why are you creating a line from the centre to the midpoint? You have already used that the two points need to be on the circle and there are an infinite number of circles satisfying this. You cannot squeeze more information out of those two points. You need to use the third requirement.
Oh, right. I just realized that- sort of like writing a third KVL equation which is the same. To use the third condition I'd have to put the radius=distance of line from center which is lengthy and prone to mistakes in an exam. I was hoping to find a shorter method but I suppose this is the only way to do it...
 
  • Like
Likes Delta2
  • #5
Orodruin said:
Also note that your equation in bold here is not a new equation. You can get it by just using your previous two so there is no new information. Your equation system therefore does not have a unique solution (two equations for three variables) and you need to use the extra information provided.
Thank you very much for your help :D
 
  • #6
$$(x-a)^2+(y-(5-a))^2=r^2 $$ Minimise for tangent: $$2(x-a)+2(y-(5-a))\frac{dy}{dx}=0 $$ with $$ y=3-3x \Rightarrow \frac{dy}{dx}=-3$$ Interesting problem - you end up with two values of a and hence two circles satisfying the given conditions.
 
  • #7
neilparker62 said:
$$(x-a)^2+(y-(5-a))^2=r^2 $$ Minimise for tangent: $$2(x-a)+2(y-(5-a))\frac{dy}{dx}=0 $$ with $$ y=3-3x \Rightarrow \frac{dy}{dx}=-3$$ Interesting problem - you end up with two values of a and hence two circles satisfying the given conditions.
how did you obtain coordinates of center as (a,5-a)?
 
  • #8
Krushnaraj Pandya said:
how did you obtain coordinates of center as (a,5-a)?
From equation of perpendicular bisector of the line drawn between the two given points.
 
  • Like
Likes SammyS
  • #9
neilparker62 said:
From equation of perpendicular bisector of the line drawn between the two given points.
ohh, really good solution- how did you think of this?
 
  • #10
Other approaches seemed to be heading for complications so I tried to keep it simple! I wasn't quite sure how to use the information about the tangent line until I realized the problem was essentially one of minimising distance between point (the circle centre) and line (y=3-3x).
 
  • #11
neilparker62 said:
Other approaches seemed to be heading for complications so I tried to keep it simple! I wasn't quite sure how to use the information about the tangent line until I realized the problem was essentially one of minimising distance between point (the circle centre) and line (y=3-3x).
Great! Thank you very much for your help.
 
  • #12
Krushnaraj Pandya said:
ohh, really good solution- how did you think of this?
A pleasure. Thanks for your kind compliment - the problem was certainly a little different from 'run of the mill' exercises in analytic geometry.
 

Related to Equation of a circle from given conditions

What is the equation of a circle?

The equation of a circle is a mathematical representation that describes all the points in a two-dimensional plane that are equidistant from a fixed point, known as the center of the circle.

How is the equation of a circle written?

The standard form of the equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) represents the coordinates of the center of the circle and r represents the radius.

What are some given conditions for the equation of a circle?

Some common given conditions for the equation of a circle include the center coordinates and the radius, the endpoints of the diameter, and the points on the circumference of the circle.

Can the equation of a circle be written in other forms?

Yes, the equation of a circle can also be written in general form as x^2 + y^2 + Dx + Ey + F = 0, where D, E, and F are constants. It can also be written in parametric form as x = h + rcos(t) and y = k + rsin(t), where (h,k) are the center coordinates and t is the angle in radians.

How can the equation of a circle be used in real life?

The equation of a circle has many practical applications in fields such as engineering, architecture, and physics. It can be used to design circular structures, calculate the path of a moving object in circular motion, and determine the location of points in a plane.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
15
Views
3K
  • Precalculus Mathematics Homework Help
Replies
16
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
1K
  • Precalculus Mathematics Homework Help
Replies
14
Views
538
  • Precalculus Mathematics Homework Help
Replies
8
Views
2K
  • Precalculus Mathematics Homework Help
Replies
21
Views
905
  • Precalculus Mathematics Homework Help
2
Replies
48
Views
4K
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
  • Precalculus Mathematics Homework Help
Replies
5
Views
2K
Back
Top