How do you graph #x^2+y^2-49=0#?

Question

Wyatt Anderson · Accepted Answer

See answer below.

We compare this to the equation of a circle. #(x-a)^2+(y-b)^2=r^2# This is a circle, center #(a,b)# and radius #r# In the equation #x^2+y^2=49=7^2# This is the equation of a circle, center #(0,0)# and radius #=7# Draw this circle, center at the origin and radius #=7# graph{x^2+y^2=49 [-18.02, 18.01, -9.01, 9.01]}

Wyatt Anderson · Answer

undefined To graph the equation $x^2 + y^2 - 49 = 0$, you can follow these steps:

1. Rewrite the equation in the standard form of a circle: $x^2 + y^2 = r^2$, where $r$ is the radius of the circle.
   
2. Compare the given equation $x^2 + y^2 - 49 = 0$ with the standard form. You can see that $r^2 = 49$, so $r = \sqrt{49} = 7$.
   
3. Identify the center of the circle. In this case, since there are no terms with $x$ or $y$ that are not squared, the center is at the origin $(0,0)$.
   
4. Plot the center point at $(0,0)$.
   
5. Use the radius $r = 7$ to plot points on the circle. You can choose various values for $x$ (e.g., $x = -7, 0, 7$) and solve for $y$ to find corresponding points on the circle.
   
6. Once you have a few points on the circle, draw a smooth curve connecting them to complete the graph.

So, the graph of $x^2 + y^2 - 49 = 0$ is a circle centered at the origin $(0,0)$ with a radius of $7$.