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

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Answer 1 · 2016-06-06T16:11:43

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

Complete the square to identify relevant features in this graph.

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

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

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