Find the equation of the circle. Find the equation of the circle whose radius is $5$ which touches the circle $x^2 + y^2 - 2x -4y - 20 = 0$ externally at the point $(5,5)$
 A: Hint:
From the fact that the circles touch eachother externally in $(5,5)$ it follows that $(5,5)$ is on the line segment that connects the centers of the circle.
A: The circle: $x^2+y^2-2x-4y-20=0$ has center $(1, 2)$ & a radius $=\sqrt{(-1)^2+(-2)^2-(-20)}=5$ & the unknown circle has a radius $5$ Hence the point $(5, 5)$ is the mid point of line joining their centers
Let the center of unknown circle be $(a, b)$ then the point $(5, 5)$ is mid point of lines joining the centers $(a, b)$ & $(1, 2)$ hence we have $$\left(\frac{a+1}{2}, \frac{b+2}{2}\right)\equiv(5, 5)$$ by comparing the corresponding coordinates we get $$\frac{a+1}{2}=5\implies a=9$$ $$\frac{b+2}{2}=5\implies b=8$$ Hence the equation of the circle having center $(9, 8)$ & a radius $5$ is given as $$(x-9)^2+(y-8)^2=5^2=25$$
$$\color{blue}{(x-9)^2+(y-8)^2=25}$$
A: HINT:
$(x-a)^2+(y-b)^2=5^2$ will touch  $(x-1)^2+(y-2)^2=5^2$
iff $5+5=\sqrt{(a-1)^2+(b-2)^2}$
Again, $(a,b), (1,2), (5,5)$ are collinear.
So, we have two equations with two unknowns 
A: Express that the circle is through the point $(5,5)$
$$(5-x_c)^2+(5-y_c)^2=5^2,$$
and that the gradients are collinear a this point
$$(x_c-5)(2\cdot5-4)-(y_c-5)(2\cdot 5-2)=0.$$
From the second equation,
$$y_c=\frac{3x_c+5}4,$$
and plugging in the first,
$$x_c^2-10x_c+9=0.$$
The solutions are $(x_c,y_c)=(1,2)$ and $(9,8)$ and we choose the second.
$$\color{green}{(x-9)^2+(y-8)^2=5^5}.$$
A: Thank you for the great help guys. Let me share you my solution for this number.
I have to get first the center of $x^2+y^2-2x-4y-20=0$
$$(x-1)^2+(y-2)^2=5^2$$
It follows that the center is $(1,2)$.
I will get the distance of the center to $(5,5)$. 
$d=5$ means radius is $5$, the same with the radius of another external circle.
So, I have $P_1 = (1,2)$, $P_2 = (5,5)$, $P_3 = (x,y)$.
I can use median formula for this to get $P_3$, given that $(5,5)$ will be the median. 
I got $P_3 = (9,8)$. 
Now I have the center of the other circle which is $(9,8)$, I make a standard form formula for circle.
$(h,k) = (9,8)$
Radius $5$ units
STANDARD FORM
$$(x-9)^2 + (y-8)^2 = 5^2$$ 
GENERAL FORM
$$x^2+y^2-18x-16y+120= 0$$ 
Thank you very much! :)
