Question

consider the circle centered at the origin with radius 3.
(a) give the equation of the circle.
(b) for each point, decide whether or not it is on the circle.

Explanation:

Step1: Write the standard equation of a circle

The standard equation of a circle with center \((h,k)\) and radius \(r\) is \((x - h)^2+(y - k)^2=r^2\). Here, \(h = 0,k = 0,r = 3\). So the equation is \(x^{2}+y^{2}=9\).

Step2: Check if the points are on the circle

• For the point \((0,-2)\):

Substitute \(x = 0,y=-2\) into \(x^{2}+y^{2}\). We get \(0^{2}+(-2)^{2}=4
eq9\).

• For the point \((3,0)\):

Substitute \(x = 3,y = 0\) into \(x^{2}+y^{2}\). We get \(3^{2}+0^{2}=9\).

• For the point \((-2,\sqrt{5})\):

Substitute \(x=-2,y = \sqrt{5}\) into \(x^{2}+y^{2}\). We get \((-2)^{2}+(\sqrt{5})^{2}=4 + 5=9\).

• For the point \((\sqrt{5},2)\):

Substitute \(x=\sqrt{5},y = 2\) into \(x^{2}+y^{2}\). We get \((\sqrt{5})^{2}+2^{2}=5+4=9\).

Answer:

(a) \(x^{2}+y^{2}=9\)
(b)

• \((0,-2)\): No
• \((3,0)\): Yes
• \((-2,\sqrt{5})\): Yes
• \((\sqrt{5},2)\): Yes