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?

Answer 1

points (0, 11) and (0, -5)

Explanation:

The equation of a circle is: (x - x0)^2 + (y - y0)^2 = r^2

where x0 is the x-coordinate of the center of the circle, y0 is the y-coordinate of the center of the circle and r is its radius. In this problem x0 = 6, y0 = 3 and r = 10 (the distance from the center of the circle to the edge). A plot of the circle is shown in the figure attached. There we can see that points (0, 11) and (0, -5) are, at the same time, on the y-axis and on the edge of the circle.

Answer 2

Possible point on y-axis are ( 0 , -5 ) and ( 0 , 11 )

Explanation:

We are given coordinate of the center of the circle ( 6 , 3 ).

The Distance from the center of the circle to the edge = 10 units.

To find: Coordinate of the point on y-axis on the edge of the circle.

Given Distance is the Length of the radius of the circle.

Radius = 10 units

We know that standard form of the coordinate of the point on the y-axis.

Coordinate of the point on the y-axis = ( 0 , y )

Also, the distance formula of the two point is given as follows,

`Distance=√((x_2-x_1)^2+(y_2-y_1)^2)`

Now, using distance formula

we have

`√((6-0)^2+(3-y)^2)=10`

`√(36+3^2+y^2-6y)=10`

`y^2-6y+45=100`

`y^2-6y-55=0`

`y^2-11y+5y-55=0`

y ( y - 11 ) + 5 ( y - 11 ) = 0

( y - 11 )( y + 5 ) = 0

y - 11 = 0 ⇒ y = 11

y + 5 = 0 ⇒ y = -5

Therefore, Possible point on y-axis are ( 0 , -5 ) and ( 0 , 11 )