Answer :
Answer:
[tex](x-2)^2+y^2=20[/tex]
Step-by-step explanation:
We want to find the equation of a circle P, and we know that the square RSTU is inscribed in the circle, this means that if we replace the points:
R(0,4) S(6,2) T(4,-4) and U(-2,-2) the equation must be verified.
Then, we have to replace the points in each option to see which one is the correct answer.
Option A: [tex](x-2)^2+y^2=68[/tex]
- R(0,4):
[tex](x-2)^2+y^2=68\\(0-2)^2+4^2=68\\(-2)^2+16=68\\4+16=68\\20\neq 68[/tex]
We can see that R doesn't belong to the equation of the circle.
We could replace the rest of the points in the equation, but since we already know that R does not belong to the circle, we already know that it is not the right option.
Option B: [tex](x+2)^2+y^2=20[/tex]
- R(0,4):
[tex](x+2)^2+y^2=20\\(0+2)^2+4^2=20\\2^2+4^2=20\\4+16=20\\20=20[/tex]
The point R(0,4) verifies the equation.
- S(6,2):
[tex](x+2)^2+y^2=20\\(6+2)^2+2^2=20\\8^2+4^2=20\\64+16=20\\80\neq 20[/tex]
The point S doesn't belong to the equation of the circle, then we already know that this isn't the correct option.
Option C: [tex](x-2)^2+y^2=20[/tex]
- R(0,4):
[tex](x-2)^2+y^2=20\\(0-2)^2+4^2=20\\(-2)^2+16=20\\4+16=20\\20=20[/tex]
The point R verifies the equation.
- S(6,2):
[tex](x-2)^2+y^2=20\\(6-2)^2+2^2=20\\4^2+4=20\\16+4=20\\20=20[/tex]
The point S verifies the equation.
- T(4,-4):
[tex](x-2)^2+y^2=20\\(4-2)^2+(-4)^2=20\\2^2+16=20\\4+16=20\\20=20[/tex]
The point T verifies the equation.
- U(-2,-2):
[tex](x-2)^2+y^2=20\\(-2-2)^2+(-2)^2=20\\(-4)^2+4=20\\16+4=20\\20=20[/tex]
The point U verifies the equation.
We can see that the four points RSTU verifies the equation. This means that the square RSTU is inscribed in circle P whose equation is: [tex](x-2)^2+y^2=20[/tex]
We can see that the graph of the equation verifies the answer.
