Question 166147
Find the radius of a circle with center at (4,1) if a chord of length 4 times square root of 2 is bisected at (7,4).
<pre><font size = 4 color = "indigo"><b>
Let O be the center of the circle (4,1).
Let P be the point (7,4), 
Let A and B be the endpoints of the chord,
so the chord is the segment AB. Draw the
graph:

{{{drawing(400,400,-3,12,-8,7,

locate(5,6.5,A), locate(9.2,2.5,B), locate(4,1,O), 
locate(7,4.6,P),

graph(400,400,-3,12,-8,7),

circle(4,1,sqrt(26)), line (5,6,9,2), 

line(3.9,1,4.1,1), line(4,.9,4,1.1),

line(6.9,4,7.1,4), line(7,3.9,7,4.1)

)}}}

Draw in OA and OP:

{{{drawing(400,400,-3,12,-8,7,

locate(5,6.5,A), locate(9.2,2.5,B), locate(4,1,O), 
locate(7,4.6,P),
triangle(4,1,7,4,5,6),
graph(400,400,-3,12,-8,7),

circle(4,1,sqrt(26)), line (5,6,9,2), 

line(3.9,1,4.1,1), line(4,.9,4,1.1),

line(6.9,4,7.1,4), line(7,3.9,7,4.1)

)}}}

We need to find the length of OA since it 
is a radius of the circle.

Triangle OPA is a right triangle, since if a
bisector of a chord passes through the center of
a circle, then it is perpendicular to the chord.

We are given that chord AB has length {{{4sqrt(2)}}},
and since P bisects it, then AP is half that length
or {{{2sqrt(2)}}}.

Next we will find the length of OP by use of the
distance formula:

{{{d=sqrt( (x[2]-x[1])^2 + (y[2]-y[1])^2 )}}}

using the given coordinates of O(4,1) and P(7,4):

{{{OP=sqrt( (7-4)^2 + (4-1)^2 )}}}

{{{OP=sqrt( (3)^2 + (3)^2 )}}}

{{{OP=sqrt(9 + 9 )}}}

{{{OP=sqrt(18)}}}

{{{OP=3sqrt(2)}}}

Now by the Pythagorean theorem,

{{{OA^2=AP^2+OP^2}}}

{{{OA^2=(2sqrt(2))^2+(3sqrt(2))^2}}}

{{{OA^2=(4*2)+(9*2)}}}

{{{OA^2=8+18}}}

{{{OA^2=26}}}

{{{OA=sqrt(26)}}}

And since OA is a radius, the circle has 
radius {{{sqrt(26)}}}.

Edwin</pre>