Question

each point on the edge of a circle is equidistant from the center of the circle. the center of a circle is located at $(6,3)$. which point on the $y$-axis could be on the edge of the circle if the distance from the center of the circle to the edge is 10 units? $(0,5)$ $(0, - 1)$ $(0,1)$ $(0, - 5)$

Explanation:

Step1: Recall the distance formula

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). The center of the circle is \((x_1,y_1)=(6,3)\), and a point on the \(y -\)axis is \((x_2,y_2)=(0,y)\). Given \(d = 10\).

Step2: Substitute into the distance formula

Substitute into \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), we get \(10=\sqrt{(0 - 6)^2+(y - 3)^2}\).

Step3: Square both sides of the equation

\(100=(0 - 6)^2+(y - 3)^2\). Since \((0 - 6)^2=36\), the equation becomes \(100 = 36+(y - 3)^2\).

Step4: Solve for \((y - 3)^2\)

Subtract 36 from both sides: \((y - 3)^2=100 - 36=64\).

Step5: Solve for \(y\)

Take the square root of both sides: \(y-3=\pm8\).

• Case 1: \(y-3 = 8\), then \(y=8 + 3=11\).
• Case 2: \(y-3=-8\), then \(y=-8 + 3=-5\).

Answer:

\((0,-5)\)