Question 1202298
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Find the radius of a circle with center at (4,1) 
if a chord of length 4√2 is bisected at (7,4)
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<pre>
Make a sketch.
In your sketch, find a right-angled triangle, which has the vertices OXY, where
O  is the center of the circle;  X  is the midpoint of the chord and 
Y  is  one of the two intersection points of the chord and the circle.


The leg XY of this triangle has the length {{{2*sqrt(2)}}}, half of the leg
of the chord; the leg OX is the vector with coordinates (3,3) = (7-4,4-1).
The length of this leg OX is  {{{3*sqrt(2)}}}.


The radius "r" of the circle is the hypotenuse of this triangle

    {{{r^2}}} = {{{(2*sqrt(2))^2}}} + {{{(3*sqrt(2))^2}}} = 4*2 + 9*2 = 8 + 18 = 26.


<U>ANSWER</U>.  The radius of the circle is  r = {{{sqrt(26)}}} units.
</pre>

Solved.