Question 931562
Complete the square to find the equation of the circle.
{{{x^2+4x+y^2-6y-12=0}}}
{{{(x^2+4x+4)+(y^2-6y+9)-12=4+9}}}
{{{(x+2)^2+(y-3)^2=25}}}
So the circle is centered at ({{{-2}}},{{{3}}}) with a radius of {{{5}}}.
The maximumum distance would be from ({{{3}}},{{{7}}}) through the center plus the radius of the circle.
First calculate the distance between ({{{3}}},{{{7}}}) and ({{{-2}}},{{{3}}}).
{{{D^2=(3-(-2))^2+(7-3)^2}}}
{{{D^2=5^2+4^2}}}
{{{D^2=25+16}}}
{{{D^2=41}}}
{{{D=sqrt(41)}}}
So then,
{{{D[max]=D+5}}}
{{{D[max]=5+sqrt(41)}}}