Question

consider the circle centered at the origin and passing through the point (9, 0).
(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 the origin \((0,0)\) is \(x^{2}+y^{2}=r^{2}\), where \(r\) is the radius.
Since the circle passes through the point \((9,0)\), substitute \(x = 9\) and \(y = 0\) into the equation: \(9^{2}+0^{2}=r^{2}\), so \(r^{2}=81\).

Step2: Write the equation of the circle

The equation of the circle is \(x^{2}+y^{2}=81\).

Step3: Check if the points are on the circle

• For the point \((8,\sqrt{17})\):

Substitute \(x = 8\) and \(y=\sqrt{17}\) into \(x^{2}+y^{2}\): \(8^{2}+(\sqrt{17})^{2}=64 + 17=81\). So this point is on the circle.

• For the point \((0,3)\):

Substitute \(x = 0\) and \(y = 3\) into \(x^{2}+y^{2}\): \(0^{2}+3^{2}=9
eq81\). So this point is not on the circle.

• For the point \((\sqrt{17},-8)\):

Substitute \(x=\sqrt{17}\) and \(y=-8\) into \(x^{2}+y^{2}\): \((\sqrt{17})^{2}+(-8)^{2}=17 + 64=81\). So this point is on the circle.

• For the point \((-5,0)\):

Substitute \(x=-5\) and \(y = 0\) into \(x^{2}+y^{2}\): \((-5)^{2}+0^{2}=25
eq81\). So this point is not on the circle.

Answer:

(a) The equation of the circle is \(x^{2}+y^{2}=81\).
(b)

• For \((8,\sqrt{17})\): Yes
• For \((0,3)\): No
• For \((\sqrt{17},-8)\): Yes
• For \((-5,0)\): No