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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)
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Respuesta :

absor201

Answer:

Option B R(2,4) is correct

Step-by-step explanation:

The equation of the circle is:

[tex](x-a)^2 + (y-b)^2 = r^2[/tex]

Where r = radius

a and b are coordinates of the center of circle.

To check which point lies on a circle, we need to verify the equation

[tex](x-6)^2 + (y-1)^2 = (5)^2[/tex]

We will check for each option.

Option A Q(1,11)

x=1 and y =11

[tex](1-6)^2 + (11-1)^2 = 25\\(-5)^2 + (10)^2 = 25\\25 + 100 = 25\\125 \neq 25[/tex]

So, Option A is incorrect

Option B R(2,4)

x =2 and y = 4

[tex](2-6)^2 + (4-1)^2 = 25\\(-4)^2 + (3)^2 = 25\\16 + 9 = 25\\25 = 25[/tex]

Option B is correct.

Option C S(4,-4)

x =4 and y =-4

[tex](4-6)^2 + (-4-1)^2 = 25\\(-2)^2 + (-5)^2 = 25\\4 + 25 = 25\\29 \neq 25[/tex]

Option C is incorrect

Option D T(9,-2)

x =9 and y =-2

[tex](9-6)^2 + (-2-1)^2 = 25\\(3)^2 + (-3)^2 = 25\\9 + 9 = 25\\18 \neq 25[/tex]

Option D is incorrect.

botellok

Answer:

B.

Step-by-step explanation:

The general equation of a circle is [tex](x-h)^{2}+(y-k)^{2} = r^{2}[/tex] where (h,k) is the center and r the radius. In this case, the general equation of the circle with radius 5 and center at (6,1) is [tex](x-6)^{2}+(y-1)^{2} = 5^{2}[/tex], so the point that satisfies the equation will be in the circle.

A.  [tex](1-6)^{2}+(11-1)^{2} = 25+100 = 125[/tex] this option is not correct.

B.   [tex](2-6)^{2}+(4-1)^{2} = 16+9= 25[/tex] this option is correct so is the answer.

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