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I'm working through the following question:

Show that if $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ are at opposite ends of a diameter of a circle then the equation of the circle is:

$$(x − x_{1})(x − x_{2}) + (y − y_{1})(y − y_{2}) = 0$$

Source: http://www.personal.soton.ac.uk/jav/soton/HELM/workbooks/workbook_2/2_6_circle.pdf

I was thinking that for the points to be at opposite ends of the diameter of a circle, the following must be true:

$$(x - x_{1})^{2} + (y - y_{1})^{2} = R^{2} = (x - x_{2})^{2} + (y - y_{2})^{2}$$

However, my expression above expands to:

$$\begin{equation*}- 2 x x_{1} + 2 x x_{2} + x_{1}^{2} - x_{2}^{2} - 2 y y_{1} + 2 y y_{2} + y_{1}^{2} - y_{2}^{2}\end{equation*}$$

... and the expression in the question expands to:

$$\begin{equation*}x^{2} - x x_{1} - x x_{2} + x_{1} x_{2} + y^{2} - y y_{1} - y y_{2} + y_{1} y_{2}\end{equation*}$$

Where am I going wrong?

Simple answers will be preferred over complex ones.

Community
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Chris Snow
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  • Indeed, the above formula does not determine x and y sufficiently, it will only cover the perpendicular bisector joining the points. To use the fact that this is a diameter, you can use the fact that the angle in a semicircle is a right angle, or, since you have found the perpendicular bisector, you can find the intersection of this point and the line joining the two given points. That point is the center, and the radius is the distance from this point to any one of the two points. But this procedure is very long, and the answer below is brilliant. – Sarvesh Ravichandran Iyer Jul 27 '17 at 11:35

2 Answers2

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What you have done is find the perpendicular bisector of $(x_1,y_1)$ and $(x_2,y_2)$ (points that are the same distance from both).

I think the simplest way to find the circle is to note that the angle in a semicircle is a right-angle, so the triangle $(x_1,y_1)(x,y)(x_2,y_2)$ has a right-angle at $(x,y)$. Lines intersect at a right-angle when the gradients have product $-1$, so $$ \frac{y-y_1}{x-x_1} \frac{y-y_2}{x-x_2} = -1, $$ at least where $x \neq x_1$ and $x \neq x_2$, and rearranging this gives the result.

Chappers
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  • Not a right-angle ($90^{\circ}$) but $180^{\circ}$. Then the rest of what you said is correct. – dt688 Jul 27 '17 at 11:36
  • @dt688 https://www.google.co.uk/search?q=%22angle+in+a+semicircle%22 Convention is "angle in a semicircle" means the angle on the boundary. – Chappers Jul 27 '17 at 11:53
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The equation of a circle with center $O(x_0,y_0)$ and radius $R$ is: $$(x-x_0)^2+(y-y_0)^2=R^2.$$ The circle with end points of its diameter at $A(x_1,y_1)$ and $B(x_2,y_2)$ will have the center $O$ and radius $R$ as follows: $$O\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$ $$R^2=\left(x_1-\frac{x_1+x_2}{2}\right)^2+\left(y_1-\frac{y_1+y_2}{2}\right)^2.$$ Now substitute these values into the circle equation and rearrange: $$\left(x-\frac{x_1+x_2}{2}\right)^2-\left(x_1-\frac{x_1+x_2}{2}\right)^2+\left(y-\frac{y_1+y_2}{2}\right)^2-\left(y_1-\frac{y_1+y_2}{2}\right)^2=0$$ Applying $a^2-b^2=(a-b)(a+b)$ will produce the desired result.

farruhota
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