## Precalculus (10th Edition)

${{\left( x-4 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=9$
The standard equation of circle with center$(h,k)$ and radius $r$ is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ The center of the circle is $(4,-2)$ The circle is tangent to the line $x=1$. Therefore, the radius is the distance from the point $(4,-2)$ to $x=1$. The distance of point $\left( x,y \right)$ from $x=a$ is $\left| x-a \right|$ Therefore, the distance of point $(4,-2)$ from $x=1$ is $\,|4-1|\,=\,3$ Therefore, the radius of the circle is $3$. This gives: ${{\left( x-4 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=9$