Derive the general formula of the equation of a circle for the points

Notice you have to use (x,y) as the variable for the general point, and you use the 1st two parts to find the center of your circle.]In summary, the general formula for the equation of a circle using the points A(x1, y1) and C(x2, y2) as its diameter is (x - (x1+x2)/2)^2 + (y - (y1+y2)/2)^2 = ((x2-x1)^2 + (y2-y1)^2)/4. This can be derived by using the midpoint formula to find the center of the circle and then using the distance formula to find the radius, setting it equal to r and simplifying
  • #1
lionely
576
2

Homework Statement



1. (a) Find the equation of the circle with the straight line joining A(1;-1) and C(3; 4)
as diameter.
(b) Hence or otherwise, derive the general formula
(x - x1)(x - x2) + (y - y1)(y - y2)=0
of the equation of a circle for the points A(x1; y1) and C(x2; y2):

Homework Equations


The Attempt at a Solution



I did part a) but part b) is baffling me could someone give me a hint on what to do?
 
Last edited:
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  • #2
lionely said:

Homework Statement



1. (a) Find the equation of the circle with the straight line joining A(1;-1) and C(3; 4)
as diameter.
(b) Hence or otherwise, derive the general formula
(x - x1)(x - x2) + (y - y1)(y - y2)
of the equation of a circle for the points A(x1; y1) and C(x2; y2):
Something missing here? I would think that a "general formula" would be an equation, which the above is not. Is this the complete problem statement?
lionely said:

Homework Equations





The Attempt at a Solution



I did part a) but part b) is baffling me could someone give me a hint on what to do?
 
  • #3
sorry it was supposed to be equal to 0
 
  • #4
lionely said:

Homework Statement



1. (a) Find the equation of the circle with the straight line joining A(1;-1) and C(3; 4)
as diameter.
(b) Hence or otherwise, derive the general formula
(x - x1)(x - x2) + (y - y1)(y - y2)=0
of the equation of a circle for the points A(x1; y1) and C(x2; y2):

Homework Equations


The Attempt at a Solution



I did part a) but part b) is baffling me could someone give me a hint on what to do?

If you did part a, do part b the same way using the variables instead of the numbers. Then see if you can get it in the required form.

[Edit] Alternatively use this hint: Those points and the point (x,y) can be used to make an angle inscribed in a semicircle, so...
 
  • #5
Yeah I tried and took the gradient of the lines and got it. Thanks.
 
  • #6
lionely said:

Homework Statement



1. (a) Find the equation of the circle with the straight line joining A(1;-1) and C(3; 4)
as diameter.
(b) Hence or otherwise, derive the general formula
(x - x1)(x - x2) + (y - y1)(y - y2)=0
of the equation of a circle for the points A(x1; y1) and C(x2; y2):
Impossible-this is NOT the equation of a circle!

Homework Equations





The Attempt at a Solution



I did part a) but part b) is baffling me could someone give me a hint on what to do?
 
  • #7
HallsofIvy said:
Impossible-this is NOT the equation of a circle!

It looks like the solution to me

[tex] (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=x^2-(x_1+x_2)x+x_1x_2+y^2-(y_1+y_2)y+y_1y_2 \\
=(x-\frac{x_1+x_2}{2})^2-(\frac{x_1+x_2}{2})^2+x_1x_2+(y-\frac{y_1+y_2}{2})^2-(\frac{y_1+y_2}{2})^2+y_1y_2=0 \\

(x-\frac{x_1+x_2}{2})^2+(y-\frac{y_1+y_2}{2})^2= \frac{x_1^2+2x_1x_2+x_2^2}{4}-x_1x_2+\frac{y_1^2+2y_1y_2+y_2^2}{4}-y_1y_2

\\

(x-\frac{x_1+x_2}{2})^2+(y-\frac{y_1+y_2}{2})^2= \frac{x_1^2-2x_1x_2+x_2^2}{4}+\frac{y_1^2-2y_1y_2+y_2^2}{4}
\\
(x-\frac{x_1+x_2}{2})^2+(y-\frac{y_1+y_2}{2})^2= \frac{(x_1-x_2)^2+(y_1-y_2)^2}{4} [/tex]

This looks like the equation of a circle with the right center and radius.
 
  • #8
The centre of the circle is the midpoint of the diameter, at O((x1+x2)/2;(y1+y2)/2). The radius is half of the diameter: R=0.5sqrt((x2-x1)2+(y2-y1)2). Writing up the equation of a circle with these parameters, arranging, factorising and simplifying, you get the desired formula.

ehild
 
  • #9
Lionely seemed to get my hint. The equation is trivially correct if you note that if ##P=(x,y)## is a point on the circle with ##A=(x_1,y_1),B=(x_2,y_2)## then angle ##APB##, being inscribed in a semicircle, is a right angle. The equation is simply ##\vec{AP}\cdot \vec{BP}=0##.
 
  • #10
LCKurtz said:
Lionely seemed to get my hint. The equation is trivially correct if you note that if ##P=(x,y)## is a point on the circle with ##A=(x_1,y_1),B=(x_2,y_2)## then angle ##APB##, being inscribed in a semicircle, is a right angle. The equation is simply ##\vec{AP}\cdot \vec{BP}=0##.

That is the nicest solution! :smile:

ehild
 
  • #11
I didn't use vectors I took the gradient but your solution was like 2 lines mine was like 5 or more . Using vectors like that is ... awesome.
 
  • #12
I'm confused what you want in part (b). The general equation for circle in two dimensions is
[itex](x-a)^2+(y-b)^2=r^2[/itex]
and comes from using the distance formula for an arbitrary center of the circle.

If you would take your two given points, use middle-point formula, this would be your circle's center. Half the length of the segment joining your two given points has length of your radius.
Use Distance formula with the general point, (x,y), the express distance from center to this general (x,y), set it equal to r, and then derive your equation.
 

FAQ: Derive the general formula of the equation of a circle for the points

What is the general formula for the equation of a circle?

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

How do you find the center of a circle using the general formula?

The center of a circle can be found by identifying the values of h and k in the general formula. These values represent the x-coordinate and y-coordinate of the center, respectively.

What does the variable "r" represent in the general formula of a circle?

The variable "r" in the general formula represents the radius of the circle. It is a measure of the distance from the center of the circle to any point on the circumference.

Can the general formula of a circle be used for circles with any center and radius?

Yes, the general formula of a circle can be used for circles with any center and radius. It is a versatile formula that can be applied to circles of different sizes and orientations.

How does the general formula of a circle relate to the Pythagorean Theorem?

The general formula of a circle is derived from the Pythagorean Theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. By applying this theorem to a circle, we can derive the general formula using the distance formula and the Pythagorean identity.

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