Question 1084921
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the center of a circle is at (4,2) and its radius is 5. find the length of the chord which is bisected at (2,-1).
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Below I placed much simpler solution.


<pre>
1.  Make a sketch to follow my arguments.

    Let the point O = (4,2) be the center of the circle,
    and let the point A = (2,-1) bisects our chord.

    Let B  and C be the endpoints of the chord.


2.  Then {{{abs(OC)^2}}} = {{{abs(OA)^2}}} + {{{abs(AC)^2}}},   or

    For |OC| you have  {{{abs(OC)^2}}} = 25  and  for  |OA| you have 

     {{{abs(OA)^2}}} = {{{(4-2)^2 + (2-(-1))^2}}} = {{{2^2 + 3^2}}} = {{{13}}}.


    Therefore,  {{{abs(AC)^2}}} = {{{5^2 - 13}}} = 12.

    Then |AC| = {{{sqrt(12)}}} = {{{2*sqrt(3)}}},

         and  |BC| = {{{4*sqrt(3)}}}.
</pre>

<U>Answer</U>.  The length of the chord is {{{4*sqrt(3)}}}.