Question

Which point lies on a circle with a radius of 5 units and center at P(6, 1)? A. Q(1, 11) B. R(2, 4) C. S(4, -4) D. T(9, -2)

Answer 1

Given the center coordinates
`(h,k)` and the radius
`r`, the equation of the circle is

`(x-h)^2 + (y-k)^2 = r^2`

So, the equation of your circle is

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

We can plug the values from the options to see which one satisfy the equation:

A)
`(1-6)^2 + (11-1)^2 = 25 +100 \\eq 25`

B)
`(2-6)^2 + (4-1)^2 = 16 +9 = 25`

So, we already found the correct option. For the sake of completeness:

C)
`(4-6)^2 + (-4-1)^2 = 4 +25 \\eq 25`

D)
`(9-6)^2 + (-2-1)^2 = 9 +9 \\eq 25`

Answer 2

B.
`R(2, 4)`

Explanation:

The equation of a circle with center
`(h, k)` and the radius,
`r`, is

`(x-h)^2+(y-k)^2=r^2`

Here, the center of the circle is at
`P(6, 1)` and the radius of the circle is 5 units.

So,
`h=6, k=1` and
`r=5`

Therefore, the equation of the circle is

`(x-6)^2+(y-1)^2=5^2`

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

Now, the point that satisfies the equation of the circle will lie on the circle.

Let us take all the points one by one.

`Q(1, 11)`

`(1-6)^2+(11-1)^2=25\\\\(-5)^2+(10)^2=25\\\\25+100=25\\\\125=25`, which is incorrect.

So,
`Q(1, 11)` does not lie on the circle.

`R(2, 4)`

`(2-6)^2+(4-1)^2=25\\\\(-4)^2+(3)^2=25\\\\16+9=25\\\\25=25`, which is correct.

So,
`R(2, 4)` lie on the circle.

`S(4, -4)`

`(4-6)^2+(-4-1)^2=25\\\\(-2)^2+(-5)^2=25\\\\4+25=25\\\\29=25`, which is incorrect.

So,
`S(4, -4)` does not lie on the circle.

`T(9, -2)`

`(9-6)^2+(-2-1)^2=25\\\\(3)^2+(-3)^2=25\\\\9+9=25\\\\18=25`, which is incorrect.

So,
`T(9, -2)` does not lie on the circle.

Hence, the point
`R(2, 4)` lie on the circle.