equation of circle tangent to line with radius Find the equation of a circle tangent to line $3x + y - 2 = 0$ at $(-1,5)$ and with radius $\sqrt{10}$. I've no idea on how to do this.
Now as per the question the circle touches the line only at one point so it is a tangent and there can be two circles touching the given line and passing through (-1,5).

so if the given line is a tangent to the circle then the centre lies on a line perpendicular to the given line, passing through (-1,5) and at a distance of $\sqrt(10)$.



slope of line perpendicular to it is $$m1*m2=-1$$ where m1=-3 which implies m2 =1/3. 

let the line be y=(1/3)x+c. Now this passes through (-1,5) substitute and get c value

The resulting equation which is perpendicular to the given line is $$3y=x+16---(1)$$

so to find the circle equation we need to find the centre of the circle which can be found with the equation shown which is nothing but the distance formula$$(x-(-1))^2+(y-5)^2=(\sqrt10)^2   ---(2)$$

since the circle has a radius of $\sqrt(10)$ units which implies the centre is at a distance of $\sqrt(10)$ units from it.

solving (1) and (2) we get $$x=-4,2$$ and $$y=4,6$$

therefore the equations of two circles are $$(x-(-4))^2+(y-4)^2=(\sqrt10)^2$$ and $$(x-2)^2+(y-6)^2=(\sqrt10)^2$$

hope this helped.
Let the equation be $$10=(x-a)^2+(y-b)^2$$
$$\implies10=(a+1)^2+(b-5)^2\ \ \ \  (1)$$
Again like equation of circle tangent to line with radius,
$$\dfrac{3a+b-2}{\sqrt{3^2+1^2}}=\pm\sqrt{10}$$
$$3a+b-2=\pm10$$
Considering '+' sign, $$3a+b-2=10\iff b=12-3a\ \ \ \  (2)$$
Use $(1),(2)$ find $a$ and hence $b$
Similarly consider '-' sign