Answer
$(x-4)^2+(y+2)^2=9$
Work Step by Step
With its center at $(4, -2)$, the tentative equation of the circle is:
$(x-h)^2+(y-k)^2=r^2
\\(x-4)^2+(y-(-2))^2=r^2
\\(x-4)^2+(y+2)^2=r^2$
The circle is tangent to the line $x=1$.
This means that the point on the line $x=1$ that is directly to the left of the center is a point on the circle.
This point is $(1, -2)$. (refer to the attached image below)
Solve for the value of $r^2$ by substituting the x and y values of the point $(1, -2)$ into the tentative equation above to obtain:
$(x-4)^2+(y+2)^2=r^2
\\(1-4)^2+(-2+2)^2=r^2
\\(-3)^2+0^2=r^2
\\9+0=r^2
\\9=r^2$
Therefore, the equation of the circle is:
$(x-4)^2+(y+2)^2=9$
