Question # Solve the equation for this circle in StandardThe form:x^{2} - 16x + y^{2} - 10y + 73 = 0Enter radicals such as sqrt{15} as "sqrt{15}"instead of using rational exponents.

ANSWERED Solve the equation for this circle in Standard
The form:
$$x^{2}\ -\ 16x\ +\ y^{2}\ -\ 10y\ +\ 73 = 0$$
Enter radicals such as $$\sqrt{15}$$ instead of using rational exponents. 2020-11-04

Consider the equation of circle $$x^{2}\ -\ 16x\ +\ y^{2}\ -\ 10y\ +\ 73 = 0$$
To change the equation into standard form of the equation $$(x\ -\ h)^{2}\ +\ (y\ -\ k)^{2} = r^{2}$$
Grouping the terms
$$(x^{2}\ -\ 16x)\ +\ (y^{2}\ -\ 10y)\ +\ 73 = 0$$
Add and subtract 64 into $$(x^{2}\ -16x)$$
and Add and subtract 25 into $$(y^{2}\ -10y)$$
$$(x^{2}\ -16x\ +\ 64\ -\ 64)\ +\ (y^{2}\ -\ 10y\ +\ 25\ -\ 25)\ +\ 73 = 0$$
$$(x^{2}\ -\ 16x\ +\ 64)\ +\ (y^{2}\ -10y\ +\ 25)\ -\ 25\ -\ 64\ +\ 73 = 0$$
$$(x\ -\ 8)^{2}\ +\ (y\ -\ 5)^{2}\ -16 = 0$$
$$(x\ -\ 8)^{2}\ +\ (y\ -\ 5)^{2} = 16$$
Further simplify it
$$(x\ -\ 8)^{2}\ +\ (y\ -\ 5)^{2} = 4^{2}$$
Which is standard form of the circle.
Compare it by standard form of the equation $$(x\ -\ h)^{2}\ +\ (y\ -\ k)^{2} = r^{2}$$
$$h = 8,\ k = 5\text{ and } r = 4$$
Therefore, centre $$(h,\ k) = (8,\ 5)$$
And radius $$r = 4$$
List four point on thw circle $$(x\ -\ 8)^{2}\ +\ (y\ -\ 5)^{2} = 4^{2}$$ is given below
$$(8,\ 1),\ (12,\ 5),\ (8,\ 9)\text{ and }(4,\ 5)$$