Question 1020736
Three different questions.  I will help with just the second question.


This way in pure text:
x^2+y^2–16x+6y+53=0


{{{x^2+y^2–16x+6y+53=0}}}-----in case not fully rendering, x^2+y^2-16x+6y+53=0
{{{x^2-16x+y^2+6y=-53}}}
Choose correct terms to complete the squares as/if necessary.
{{{x^2-16x+64+y^2+6y+9=-53+64+9}}}
{{{(x-8)^2+(y+3)^2=73-53}}}
{{{highlight((x-8)^2+(y+3)^2=20)}}}
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Center point of the circle is  (8,-3).



How to complete the square is described and discussed here:
<a href="http://www.algebra.com/my/Completing-the-Square-to-Solve-General-Quadratic-Equation.lesson?content_action=show_dev">http://www.algebra.com/my/Completing-the-Square-to-Solve-General-Quadratic-Equation.lesson?content_action=show_dev</a>