Question 534153
The given circle equation can be written as
{{{ x^2+y^2+4x+6y=23}}}
and completing squares as
{{{ x^2+4x+4+y^2+6y+9=23+4+9}}} or {{{(x+2)^2+(y+3)^2=36}}}
showing that its radius is 6 and its center is (-2, -3).
The distance between the centers is
{{{sqrt((4-(-2))^2+(5-(-3))^2)=sqrt(6^2+8^2)=sqrt(100)=10}}}
If the circles meet in between the centers (they are externally tangent), the radius of the second circle will be
{{{10-6=4}}}
and the equation for the second circle will be
{{{(x-4)^2+(y-5)^2=16}}}
which can be written as
{{{ x^2+y^2-8x-10y+25=0}}}
If we make the second circle contain the first one (internally tangent circles), then the radius would be
{{{10+6=16}}}
and the equation for the second circle would be
{{{(x-4)^2+(y-5)^2=256}}}
which can be written as
{{{ x^2+y^2-8x-10y-215=0}}}