## College Algebra (10th Edition)

$(x-4)^2+(y+2)^2=9$
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: $(x-4)^2+(y+2)^2=9$