Question 1071413
Assume that the center of the circles is the origin (0,0),
The outer circle is,
{{{x^2+y^2=6^2}}}
1.{{{x^2+y^2=36}}}
If we make a circle centered at the intersection of the 3 and 4 lines with a radius of 4, it would be,
{{{x^2+(y-3)^2=4^2}}}
2.{{{x^2+(y+3)^2=16}}}
We can find the intersection of the outer circle and the new circle,
From 1,
{{{x^2=36-y^2}}}
Substituting into 2,
{{{36-y^2+(y+3)^2=16}}}
{{{36-y^2+y^2+6y+9=16}}}
{{{6y+45=16}}}
{{{6y=-29}}}
{{{y=-29/6}}}
So then,
{{{x^2+(-29/6)^2=36}}}
{{{x^2=36-(-29/6)^2}}}
{{{x^2=1296/36-841/36}}}
{{{x^2=455/36}}}
{{{x=-sqrt(455)/6}}}
So now find the distance between ({{{0}}},{{{-6}}}) and ({{{-sqrt(455)/6}}},{{{-29/6}}}).
{{{D^2=(-sqrt(455)/6-0)^2+(-29/6-(-6))^2}}}
{{{D^2=455/36+(-29/6+36/6)^2}}}
{{{D^2=455/36+(7/6)^2}}}
{{{D^2=455/36+49/36}}}
{{{D^2=504/36}}}
{{{D^2=14}}}
{{{D=sqrt(14)}}}

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*[illustration crt3.JPG].