Question

b. a circle with equation x² + y² - 6x + 16y + 57 = 0

Explanation:

Step1: Complete the square for x - terms

Group the x - terms and y - terms: $(x^{2}-6x)+(y^{2}+16y)+57 = 0$.
For the x - terms, $x^{2}-6x=(x - 3)^{2}-9$ (using the formula $(a - b)^2=a^{2}-2ab + b^{2}$, here $a = x$ and $b = 3$).

Step2: Complete the square for y - terms

For the y - terms, $y^{2}+16y=(y + 8)^{2}-64$ (using the formula $(a + b)^2=a^{2}+2ab + b^{2}$, here $a = y$ and $b = 8$).

Step3: Rewrite the equation

Substitute the completed - square expressions back into the original equation:
$(x - 3)^{2}-9+(y + 8)^{2}-64+57 = 0$.
Simplify the equation: $(x - 3)^{2}+(y + 8)^{2}-9 - 64+57 = 0$.
$(x - 3)^{2}+(y + 8)^{2}-16 = 0$.
$(x - 3)^{2}+(y + 8)^{2}=16$.

Answer:

The standard - form equation of the circle is $(x - 3)^{2}+(y + 8)^{2}=16$, with center $(3,-8)$ and radius $r = 4$.