Question

consider the circle centered at the origin and passing through the point $(0,8)$.
(a) give the equation of the circle.
(b) for each point, decide whether or not it is on the circle.

Explanation:

Step1: Find the radius of the circle

The standard equation of a circle centered at \((h,k)\) is \((x - h)^2+(y - k)^2=r^2\). Since the circle is centered at the origin \((0,0)\), \(h = 0,k = 0\). The circle passes through \((0,8)\), so the radius \(r=\sqrt{(0 - 0)^2+(8 - 0)^2}=8\).

Step2: Write the equation of the circle

Substitute \(h = 0,k = 0,r = 8\) into the standard equation \((x - h)^2+(y - k)^2=r^2\), we get \(x^{2}+y^{2}=64\).

Step3: Check if the points are on the circle

• For the point \((\sqrt{15},- 4)\):

Substitute \(x=\sqrt{15},y = - 4\) into \(x^{2}+y^{2}\), we have \((\sqrt{15})^{2}+(-4)^{2}=15 + 16=31
eq64\).

• For the point \((7,0)\):

Substitute \(x = 7,y = 0\) into \(x^{2}+y^{2}\), we have \(7^{2}+0^{2}=49
eq64\).

• For the point \((0,-8)\):

Substitute \(x = 0,y=-8\) into \(x^{2}+y^{2}\), we have \(0^{2}+(-8)^{2}=64\).

• For the point \((6,\sqrt{7})\):

Substitute \(x = 6,y=\sqrt{7}\) into \(x^{2}+y^{2}\), we have \(6^{2}+(\sqrt{7})^{2}=36 + 7=43
eq64\).

Answer:

(a) The equation of the circle is \(x^{2}+y^{2}=64\).
(b)

\((x,y)\)	Yes	No
\((7,0)\)		\(○\)
\((0,-8)\)	\(○\)
\((6,\sqrt{7})\)		\(○\)