Question

Which point lies on a circle with a radius of 5 units and center at P(6, 1)?

Answer 1

If you want just some random point on the circle it could be (6,6)

Answer 2

One point could be (1,1)

Explanation:

1. The equation of the circle is
`(x-h)^2 +(y-k)^2=r^2`

h and k are the (x,y) coordinates of the center of the circle respectively. In this sense, h=6 and k=1, and r is the radius of the circle which is 5. Replacing we get

`(x-6)^2+(y-1)^2=25`

2. The x-coordiante of the center of the circle is at x=6. The radius is 5, therefore, the left end of the circle is at x=1 and the right end is at x=11. This means that the circle can take x-coordiantes from 1 to 11, and for each we will have a Y-coordinate.

3. To get the Y-coordinate we isolate Y from the circle equation.

`(x-6)^2+(y-1)^2=25\\(y-1)^2=25-(x-6)^2\\y-1=√(25-(x-6)^2) \\y=√(25-(x-6)^2)+1`

4. From the last equation, we can repalce x by any number between 1 and 11, and we will get the respective Y-coordinate, for example:

If we take x=1, replacing we get:

`y=√(25-(1-6)^2 )+1\\y=√(25-(-5)^2 )+1\\\\y=√(25-25 )+1\\\\y=\ 0+1\\\\\\y=1`

So, one point could be (1,1)

5. If you want to get another point, you could replace X by 2,3,4,5,6,7,8,9,10 and 11 in the previous equation, and for each one you will get a new Y-coordinate.