Equation of a circle with centre (2, 1) and which cuts off a chord of length 4 on the line $3x+4y-5=0$  is 

d=perpendicular distance from centre (2,1) = ($x_1,y_1$) of the circle to the chord 3x+4y-5=0

d=$\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

$d=\frac{\left|6+4-5\right|}{\sqrt{25}}=1$

we have length of the chord is $4=2\sqrt{r^2-d^2}$

$2=\sqrt{r^2-d^2}$

4=$r^2-d^2$

$r=\sqrt{d^2+4}=\sqrt{5}$

$\therefore {\left(x-2\right)}^2+{\left(y-1\right)}^2=5$

$\Rightarrow x^2+y^2-4x-2y=0$