Hello, cooltowns!
Two circles touch the $\displaystyle y$axis and pass through the points $\displaystyle Q(1, 0)$ and $\displaystyle P(5, 2).$
Find the equations of the two circles and determine their radii and centers.
We want to locate the center of the circle $\displaystyle C(x,y)$
. . which is equidistant from $\displaystyle P, Q$ and the $\displaystyle y$axis. Code:

 C
Y o * * * o (x,y)
 * *
 * *
 Q * *
  +  o      *    
 (1,0) o P
 (5,2)
We want: .$\displaystyle CP \,=\,CQ \,=\,CY$
$\displaystyle \text{Then: }\;\underbrace{\sqrt{(x5)^2 + (y2)^2}}_{[1]} \;=\;\underbrace{\sqrt{(x1)^2 + y^2}}_{[2]} \;=\;\underbrace{x}_{[3]}$
Square [2] and [3]: .$\displaystyle (x1)^2 + y^2 \:=\:x^2 \quad\Rightarrow\quad x\:=\:\frac{y^2+1}{2}$ .(a)
Square [1] and [3]: .$\displaystyle (x5)^2 + (y2)^2 \:=\:x^2 \quad\Rightarrow\quad y^2 + 4y + 29 \:=\:10x$ .(b)
Substitute (a) into (b): .$\displaystyle y^2 + 4y + 29 \:=\:10\left(\frac{y^2+1}{2}\right) \quad\Rightarrow\quad y^2y6\:=\:0 $
Then: .$\displaystyle (y3)(y+2)\:=\:0 \quad\Rightarrow\quad y \:=\:3,\:\text{}2 \quad\Rightarrow\quad x \:=\:5,\:\tfrac{5}{2}$
The circles are: .$\displaystyle \begin{Bmatrix}(x5)^2 + (y3)^2 \:=\:25 & C(5,3) & r = 5 \\ \\[3mm]
\left(x\frac{5}{2}\right)^2 + (y+2)^2 \:=\:\frac{25}{4} & C\left(\frac{5}{2},\:\text{}2\right) & r = \frac{5}{2} \end{Bmatrix} $