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
[tex](x-6)^2+(y-1)^2=5^2[/tex]
[tex](x-6)^2+(y-1)^2=25[/tex]
If Q(1,11) lies on this circle, then it must satisfy its equation.
[tex](1-6)^2+(11-1)^2=25[/tex]
[tex](-5)^2+(10)^2=25[/tex]
[tex]25+100=25[/tex], this statement is false.
If R(2,4) lies on this circle, then it must satisfy its equation.
[tex](2-6)^2+(4-1)^2=25[/tex]
[tex](-4)^2+(3)^2=25[/tex]
[tex]16+9=25[/tex], this statement is false.
If S(4,-4) lies on this circle, then it must satisfy its equation.
[tex](4-6)^2+(-4-1)^2=25[/tex]
[tex](-2)^2+(-5)^2=25[/tex]
[tex]4+25=25[/tex], this statement is false.
If T(9,-2) lies on this circle, then it must satisfy its equation.
[tex](9-6)^2+(-2-1)^2=25[/tex]
[tex](4)^2+(-3)^2=25[/tex]
[tex]16+9=25[/tex], this statement is TRUE.
The correct answer is D
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 =[tex]\sqrt{(1-6)^2+(11-1)^2} =\sqrt{125} \\=5\sqrt{3}[/tex]
PR = [tex]\sqrt{(2-6)^2+(4-1)^2} =\sqrt{25} =5[/tex]
PS =[tex]\sqrt{(4-6)^2+(-4-1)^2} \\=\sqrt{29}[/tex]
PT =[tex]\sqrt{(9-6)^2+(-2-1)^2} \\=\sqrt{18} \\=3\sqrt{2}[/tex]
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)
