Answer:
The correct answer is C
Step-by-step explanation:
The given circle is centered at A(3, 1) and passes through the origin (0, 0).
The radius of this circle can be found using the distance formula;
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex].
[tex]\Rightarrow r=\sqrt{(3-0)^2+(1-0)^2}[/tex].
[tex]\Rightarrow r=\sqrt{(3)^2+(1)^2}[/tex].
[tex]\Rightarrow r=\sqrt{9+1}[/tex].
[tex]\Rightarrow r=\sqrt{10}[/tex].
The equation of this circle is given by
[tex](x-a)^2+(y-b)^2=r^2[/tex]
Where [tex](a,b)[/tex] is the centre.
We substitute the center and radius to obtain;
[tex](x-3)^2+(y-1)^2=(\sqrt{10})^2[/tex]
[tex](x-3)^2+(y-1)^2=10[/tex]
Option A
When we substitute (2,-2) into this equation we get;
[tex](2-3)^2+(-2-1)^2=10[/tex]
[tex]1+9=10[/tex]
This is true
Option B
When we substitute K(6,0), we get,
[tex](6-3)^2+(0-1)^2=10[/tex]
[tex]9+1=10[/tex]
This is also true
Option C
We substitute L(4,-4) to get;
[tex](4-3)^2+(-4-1)^2=10[/tex]
[tex]1+25=10[/tex]
This is false. Hence L(4,-4) does not lie on the circle.
Option D
We substitute M(2,4)
[tex](2-3)^2+(4-1)^2=10[/tex]
[tex]1+9=10[/tex]
This is also true.
The correct answer is C
