Answer: The required equation of the circle is [tex]x^2+y^2-14x+6y+9=0.[/tex]
Step-by-step explanation: We are given to find the equation of the circle with center at the point (7, -3) and radius of length 7 units.
We know that
the standard equation of a circle with center at the point (h, k) and radius of length r units is given by
[tex](x-h)^2+(y-k)^2=r^2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
For the given circle,
center, (h, k) = (7, -3) and radius, r = 7 units.
From equation (i), we get
[tex](x-7)^2+(y-(-3))^2=7^2\\\\\Rightarrow (x-7)^2+(y+3)^2=49\\\\\Rightarrow x^2-14x+49+y^2+6y+9=49\\\\\Rightarrow x^2+y^2-14x+6y+9=0.[/tex]
Thus, the required equation of the circle is [tex]x^2+y^2-14x+6y+9=0.[/tex]
