How do you graph $X^{2} + Y^{2} - 10 X - 4 Y + 28 = 0$ $X^2+Y^2-10X-4Y+28=0$ ?

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How do you graph $X^{2} + Y^{2} - 10 X - 4 Y + 28 = 0$ $X^2+Y^2-10X-4Y+28=0$ ?

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Answer 1 · 2016-06-06T16:11:43

A circle with centre $(5, 2)$ $(5,2)$ and a radius of $1$ $1$ unit.

Explanation:

Complete the square to identify relevant features in this graph.

$X^{2} + Y^{2} - 10 X - 4 Y + 28 = 0$ $X^2 + Y^2 - 10X - 4Y + 28 = 0$
$X^{2} - 2 (5) X + 5^{2} + Y^{2} - 2 (2) Y + 2^{2} = 5^{2} + 2^{2} - 28$ $X^2 - 2(5)X + 5^2 + Y^2 - 2(2)Y + 2^2 = 5^2 + 2^2 - 28$
${(X - 5)}^{2} + {(Y - 2)}^{2} = 1$ $(X-5)^2+(Y-2)^2=1$

Thus, we see that this graph is a circle with centre $(5, 2)$ $(5,2)$ and a radius of $1$ $1$ unit.

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How do you graph $X^{2} + Y^{2} - 10 X - 4 Y + 28 = 0$ $X^2+Y^2-10X-4Y+28=0$ ?

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