The line $ y=x/3+5$ is a tangent to a circle with centre $(-2,1)$. Find the equation of the circle

Question

I've been given this as a question for my maths AS homework but don't understand how to start. I understand that you need to find the radius to complete the equation but don't know how to do this. So far I have worked out that $(x+2)^2+(y-1)^2=r^2 $ Any Ideas of what to do next?

Answer 1

Hint: find the equation of the line perpendicular to $y=\frac13x+5$ and passing through $(-2,1)$. That line includes the radius from $(-2,1)$ to the point of tangency.

Answer 2

The hint of Téophile suggests a possible approach.

If you want continue in your approach than note that the system $$ \begin{cases} y=\frac{1}{3}x+5\\ (x-2)^2+(y-1)^2=r^2 \end{cases} $$ has only one (double) solution if the line is tangent to the circle, and this is done only if the discriminant of the second degree equation that we find substituting $y$ from the first to the second equation is null.

This discriminant is a function $\Delta(r)$ of $r$ and solving the equation $\Delta(r)=0$ you can find the value of $r$.

Answer 3

The equation of the line is $$D\;:\;\; x-3y+15=0$$

the distance to the center gives the radius.

$$R=\frac{|-2-3+15|}{\sqrt{1^2+3^2}}=1$$

the circle's equation is

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