Question 1156829
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            It is really nice problem, and it admits extraordinary beautiful solution,  unexpected,  elegant and without boring calculations.


            But my method is totally different from the instructions in your post.

            Also,  my method is different from the @greenestamps approach.


            It rarely happens to meet such beautiful problem.  I will show you the solution,

            and you can learn  A  LOT  from it . . . 



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

    +-------------------------------------------------------------------------------+
    |  Two chords in a circle are congruent (have the same length) if and only if   |
    |  the chords 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 course).


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


    Notice that the radius of the circle is 5 units, while the distance from the center (0,0)  is d= {{{sqrt(5)}}}, the lesser value,
    so the lines really intersect the circle.


Hence, the chords are equally remoted from the center, and therefore, have equal lengths.
</pre>

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The solution is completed, &nbsp;and the statement is &nbsp;PROVED.