Lesson Two chords in a circle are congruent if and only if they are equally remoted from the center

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<H2>Two chords in a circle are congruent if and only if they are equally remoted from the center</H2>

The central point of this lesson is the Problem 3 below.

Thus problem is fantastically beautiful, &nbsp;but I need to prepare you for better understanding of its solution.

So, &nbsp;the problems &nbsp;1 &nbsp;and &nbsp;2 &nbsp;serve to worm your mind.

<H3>Problem 1</H3>A chord &nbsp;6 cm long is &nbsp;2 cm from the center of a circle. 
How long is a chord that is &nbsp;1 &nbsp;cm from the center of the same circle?

<B>Solution</B>

<pre>
Make a sketch.

Draw the perpendicular radius to the first chord.

Find right angled triangle in the sketch.

For this right angled triangle, write the Pythagorean formula

r^2 = d^2 + L^2,    (1)

where r is the radius of the circle, d is the distance of the chord from the center of the circle 
and L is half the chord's length.

Then from (1) you get

r^2 = 2^2 + (6/2)^2 = 4 + 3^2 = 4 + 9 = 13.

Thus the radius of the circle is  {{{sqrt(13)}}}.

Now write equation (1) for the chord, which is 1 cm from the center of the same circle

13 = 1^2 + (x/2)^2

where x the length of this chord.

You then get

(x/2)^2 = 13 - 1 = 12;

x/2 = {{{sqrt(12)}}} = {{{2*sqrt(3)}}}

<U>ANSWER</U>.  The length of the cord is  {{{4*sqrt(3)}}} = 6.928 centimetres.
</pre>

<H3>Problem 2</H3>Find the radius of a circle in which a chord &nbsp;10 inches long is &nbsp;1 inch from the midpoint of the arc it forms?

<B>Solution</B>

<pre>
Make a sketch.

Find a right triangle in the sketch.

Let R be the unknown radius.

Then for R you have this Pythagorean equation

R^2 = {{{(10/2)^2}}} + {{{(R-1)^2}}}.

Simplify it

r^2 = 25 + R^2 - 2R + 1

2R = 26

R = 13 inches.    <U>ANSWER</U>
</pre>

<H3>Problem 3</H3>Show that the lines y = 2x − 5 and −2x + 11y = 25 create chords of equal length when they intersect the circle x2 + y2 = 25.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It is really nice problem, and it admits extraordinary beatiful solution, unexpected, elegant and without boring calculations.

<pre>
The idea of the solution is very simple:

+-------------------------------------------------------------------------------+
    |    Two cords in a circle are congruent (have the same length) if and only if  |
    |    they are equally remoted from the center of the circle.                    |
    +-------------------------------------------------------------------------------+

Indeed, if the circle has the radius R and a chord has the lenght L, then from Pythagorean theorem
you momentarily get this equation

{{{d^2}}} + {{{(L/2)^2}}} = {{{R^2}}}.    (1)

where d is the distance from the center to the chord (the minimal distance, of cource).

So, if two chords are given with the same lengths L, then from equation (1) the are equally remoted from the center.

And vice versa, if two chords are equally remoted from the center, then the formula (1) shows that in this case

their lengths will be the same.

So, this statement is proved.

Hence, the only step to do is to check that the chords, defined by the given equations, are equally remoted 
from the center of the circle.

There is a remarkable formula to calculate the distance from a given point to a given straight line in a coordinate plane.

Let the straight line in a coordinate plane is defined in terms of its linear equation

a*x + b*y + c = 0,    (2)

where "a", "b" and "c" are real numbers, and let P = ({{{x[0]}}},{{{y[0]}}}) be the point in the coordinate plane.

Then the distance from the point P to the straight line is equal to

d = {{{abs(a*x[0] + b*y[0] + c)/sqrt(a^2 + b^2)}}}.   (3)

Regarding this formula, see the lesson
    <A HREF=https://www.algebra.com/algebra/homework/Vectors/The-distance-from-a-point-to-a-straight-line-in-a-coordinate-plane.lesson>The distance from a point to a straight line in a coordinate plane</A>
in this site.

Your first straight line is  y = 2x − 5, or, in the form (2),  2x - y - 5 = 0.

So, your coefficients are a= 2, b= -1, c= -5, and the coordinates of the point are  {{{x[0]}}} = 0,  {{{y[0]}}} = 0 (the center of your circle).

Substitute this data into the formula (3) to get the distance from the center to the line

d = {{{abs(2*0 + (-1)*0 - 5)/sqrt(2^2+(-1)^2)}}} = {{{abs(-5)/sqrt(5)}}} = {{{5/sqrt(5)}}} = {{{sqrt(5)}}}.

Your second straight line is  −2x + 11y = 25, or, in the form (2),  2x - 11y + 25 = 0.

So, your coefficients are a= 2, b= -11, c= 25, and the coordinates of the point are  {{{x[0]}}} = 0,  {{{y[0]}}} = 0 (the center of your circle).

Substitute this data into the formula (3) to get the distance from the center to the second line

d = {{{abs(2*0) + (-11)*0 + 25)/sqrt(2^2+(-11)^2)}}} = {{{abs(25)/sqrt(125)}}} = {{{25/(5*sqrt(5))}}} = {{{sqrt(5)}}}.

Thus you get the same distance d = {{{sqrt(5)}}} for both lines.

It means that the chords are equally remoted from the center, and therefore, have equal lengths.

The solution is completed, and the statement is PROVED.
</pre>

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