Question

Select the correct answer. 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

Answer:

T(9,-2)

Step-by-step explanation:

The circle has radius 5 units and center P(6,1).

The equation of this circle is

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

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

If Q(1,11) lies on this circle, then it must satisfy its equation.

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

$(-5)^2+(10)^2=25$

$25+100=25$, this statement is false.

If R(2,4) lies on this circle, then it must satisfy its equation.

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

$(-4)^2+(3)^2=25$

$16+9=25$, this statement is false.

If S(4,-4) lies on this circle, then it must satisfy its equation.

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

$(-2)^2+(-5)^2=25$

$4+25=25$, this statement is false.

If T(9,-2) lies on this circle, then it must satisfy its equation.

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

$(4)^2+(-3)^2=25$

$16+9=25$, this statement is TRUE.

The correct answer is D

Answer 2

Answer:

Option B)

R(2,4)

Step-by-step explanation:

Given that P(6,1) is the center of the circle.

Let us find each point distance from P if equal to 5 units, then the point lies on the circle

PQ =$\sqrt{(1-6)^2+(11-1)^2} =\sqrt{125} \\=5\sqrt{3}$

PR = $\sqrt{(2-6)^2+(4-1)^2} =\sqrt{25} =5$

PS =$\sqrt{(4-6)^2+(-4-1)^2} \\=\sqrt{29}$

PT =$\sqrt{(9-6)^2+(-2-1)^2} \\=\sqrt{18} \\=3\sqrt{2}$

We see that only R is having a distance of 5 units from P.

Hence R lies on the circle

Option B)

R(2,4)