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) $(1-6)^2 + (11-1)^2 = 25 +100 \neq 25$

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

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

C) $(4-6)^2 + (-4-1)^2 = 4 +25 \neq 25$

D) $(9-6)^2 + (-2-1)^2 = 9 +9 \neq 25$

Answer 2

Answer:

B. $R(2, 4)$

Step-by-step 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.