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

Answer:

One point could be (1,1)

Step-by-step 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.

$(x-6)^2+(y-1)^2=25\\(y-1)^2=25-(x-6)^2\\y-1=\sqrt{25-(x-6)^2} \\y=\sqrt{25-(x-6)^2}+1$

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:

$y=\sqrt{25-(1-6)^2 }+1\\y=\sqrt{25-(-5)^2 }+1\\\\y=\sqrt{25-25 }+1\\\\y=\ 0+1\\\\\\y=1$

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.