Question 1184227

the center of a circle is at ({{{-3}}},{{{2}}}) =>{{{h=-3}}} and {{{k=2}}}

and its radius is {{{r=7}}}

equation is:

{{{(x-h)^2+(y-k)^2=r^2}}}

{{{(x-(-3))^2+(y-2)^2=7^2}}}

{{{(x+3)^2+(y-2)^2=49}}}


 Find the length of the chord, which is bisected at ({{{3}}},{{{ 1}}})



if the chord  is bisected at ({{{3}}},{{{ 1}}}), find solutions for intersection of the circle with vertical line x=3

<a href="https://ibb.co/bmMRGN6"><img src="https://i.ibb.co/bmMRGN6/circle.png" alt="circle" border="0"></a>


{{{A}}} is the {{{midpoint }}}of the chord so angle {{{CAB}}} is a {{{right}}} angle. 
Use coordinates ({{{-3}}},{{{2}}}) and ({{{3}}},{{{1}}}) to find the length of the line segment {{{CA}}}. 
Now you can use Pythagoras theorem to find the length of the line segment {{{AB}}}, which is {{{half}}} the length of the chord.

{{{CA}}}=distance between points ({{{-3}}},{{{2}}}) and ({{{3}}},{{{1}}}) 
{{{CA=sqrt((3-(-3))^2+(1-2)^2)}}}
{{{CA=sqrt(6^2+1^2)}}}
{{{CA=sqrt(37) }}}
{{{CB=7}}}

{{{(AB)^2=7^2-(sqrt(37) )^2}}}
{{{(AB)^2=49-37}}}
{{{(AB)^2=12}}}
{{{AB =sqrt(12)}}}
{{{AB =2sqrt(3)}}} -> half the length of the chord

the length of the chord will be {{{4sqrt(3) }}}=>answer