Step 1: Problem
To determine the equation of a circle with centre (5, -3) and radius 7
Step 2: Substitute the centre value and radius to the equation
[tex]\begin{gathered} \text{Equation of a circle} \\ (x-a)^2+(y-b)^2=r^2 \\ \text{centre (5,-3) , radius =7} \\ a=5,\text{ b=-3, r=7} \\ (x-5)^2+(y-(-3))^2=7^2 \end{gathered}[/tex]Step 3: Simplify the above equation
[tex]\begin{gathered} (x-5)(x-5)\text{ + (y+3)(y+3) = 49} \\ x^2-5x-5x+25+y^2+3y+3y+9=49 \\ x^2-10x+25+y^2+6y+9=49 \\ x^2+y^2-10x+6y=49-25-9 \\ x^2+y^2-10x+6y=15 \\ x^2+y^2-10x+6y-15=0 \end{gathered}[/tex]Hence the equation of the circle is
[tex]x^2+y^2-10x+6y-15=0[/tex]