How do you graph $x^{2} + y^{2} - 49 = 0$ $x^2+y^2-49=0$ ?

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Answer 1 · 2016-12-03T17:32:45

See answer below.

We compare this to the equation of a circle.

${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$

This is a circle, center $(a, b)$ $(a,b)$ and radius $r$ $r$

In the equation

$x^{2} + y^{2} = 49 = 7^{2}$ $x^2+y^2=49=7^2$

This is the equation of a circle, center $(0, 0)$ $(0,0)$ and radius $= 7$ $=7$

Draw this circle, center at the origin and radius $= 7$ $=7$

graph{x^2+y^2=49 [-18.02, 18.01, -9.01, 9.01]}

How do you graph x2+y2−49=0x^2+y^2-49=0?