Answer
$${\left( {x - \frac{1}{2}} \right)^2} + {\left( {y + \frac{3}{4}} \right)^2} = 1$$
Work Step by Step
$$\eqalign{
& 16{x^2} + 16{y^2} - 16x + 24y - 3 = 0 \cr
& {\text{Group terms}} \cr
& \left( {16{x^2} - 16x} \right) + \left( {16{y^2} + 24y} \right) = 3 \cr
& 16\left( {{x^2} - x} \right) + 16\left( {{y^2} + \frac{3}{2}y} \right) = 3 \cr
& {\text{Complete the square}} \cr
& 16\left( {{x^2} - x + \frac{1}{4}} \right) + 16\left( {{y^2} + \frac{3}{2}y + \frac{9}{{16}}} \right) = 3 + 16\left( {\frac{1}{4}} \right) + 16\left( {\frac{9}{{16}}} \right) \cr
& 16{\left( {x - \frac{1}{2}} \right)^2} + 16{\left( {y + \frac{3}{4}} \right)^2} = 16 \cr
& {\left( {x - \frac{1}{2}} \right)^2} + {\left( {y + \frac{3}{4}} \right)^2} = 1 \cr
& {\text{The equation is in the form }}{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} \cr
& {\text{Represents a circle centered at }}\left( {h,k} \right){\text{ with radius }}r \cr
& {\text{Center: }}\left( {h,k} \right) = \left( {\frac{1}{2}, - \frac{3}{4}} \right) \cr
& {\text{Radius: }}r = 1 \cr
& {\text{Graph}} \cr} $$
