# Circle equation

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.

• What did you try? – Archis Welankar Jun 27 '17 at 5:54
• The circle touches the line. That means, the line is a tangent to the circle. Moreover, a line is infinite. How will you find its "midpoint"? So, the question means to say that the given line is the tangent to the circle and the center is given. Now, you can find the perpendicular distance to the line, which will be the radius and hence proceed to finding the equation of the circle. – Aniruddha Deshmukh Jun 27 '17 at 5:56
• I agree that the phrase "touches the line" is unclear! But you can be sure that the line does not touch the center of the circle because $2 - 1 + 1 \neq 0$ (that is, the center does not satisfy the equation of the line). – manthanomen Jun 27 '17 at 6:08
• Many many thanks – Fred Jun 27 '17 at 6:55

$-x+y+1=0$ is an equation of the tangent.
$R=\frac{|-1+2+1|}{\sqrt{(-1)^2+1^2}}=\sqrt2$, which gives the answer: $(x-1)^2+(y-2)^2=2$.
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
$$(x-1)^2+(y-2)^2=2$$