Circle equation

Question

I found this questions saying

Determine the equation of the circle whose center is the point $(1,2)$ and touches the line $y-x+1=0$

Now I am not exactly sure if this means that the line touches the center or it is a tangent, but assuming it touches the center. Is there like some unique way(formula) for solving this by only knowing the midpoint (center) on the given line and using system of equation or something similar to find the two points from which I could get the radius.

If there is a much easier way please enlighten me, always had trouble dealing with circles.

Thanks in advance for replies.

Answer 1

$-x+y+1=0$ is an equation of the tangent.

$R=\frac{|-1+2+1|}{\sqrt{(-1)^2+1^2}}=\sqrt2$, which gives the answer: $(x-1)^2+(y-2)^2=2$.

Answer 2

With a quick sketch it is obvious that the line given is $L: y=x-1$ has a slope of $1$ and passes through $A(1,0)$ and $B(2,1)$.

Also, given circle centre $C(1,2)$, it is obvious that $C,B$ are opposite vertices of a single grid square giving $CB=\sqrt{2}$ with slope of $CB$ being $-1$.

Hence $CB\perp L$ and as such $B$ must be the tangential point of $L$ to the circle, and $CB$ is the radius of the circle. This gives the equation of the circle as

$$(x-1)^2+(y-2)^2=2$$