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High School
Algebra
Algebra I
Exponents and radicals

coexpennan

Answered question

2021-09-22

Rewrite the equation for this circle in Standard Form:

x^{2} - 16 x + y^{2} - 10 y + 73 = 0

Answer & Explanation

Margot Mill

Skilled2021-09-23Added 106 answers

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

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