Finding the equation of a circle. A circle of radius $2$ lies in the first quadrant touching both axis. Find the equation of the circle centered at $(6,5)$ and touching the above circle externally.
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Since the radius of the first circle is $2$ and touches both axes. It follows that $h,k$ is $(2,2)$. The formula for this circle is 
$$(x-2)^2 + (y-2)^2 = 4 .$$
Given that the center of the second circle is $(6,5)$, then we can get the distance to another circle. The distance between $(6,5)$ and $(2,2)$ is $5$. 
We can subtract and get the radius of the second circle. $r=5-2$. 
Then, the radius is $3$.
The equation of the second circle is 
$$(x-6)^2 + (y-5)^5 = 9 .$$
Thank you everyone! :)

Equation of circle of radius 2 and touching the axes is $(x-2)^2+(y-2)^2=4$ and its center is $(2,2)$.
Center of required circle is (6,5).Radius of required circle is half of distance between (6,5) and (2,2) because these two circles touch externally.Therefore radius of required circle$=r=5-2$
Now use $(x-h)^2+(y-k)^2=r^2$ and get the equation. 
