Question

Square RSTU is inscribed in circle P. Given the coordinates for the vertices of the square, find the equation of circle P.

Answer 1

Answer:

$(x-2)^2+y^2=20$

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: $(x-2)^2+y^2=68$

• R(0,4):

$(x-2)^2+y^2=68\\(0-2)^2+4^2=68\\(-2)^2+16=68\\4+16=68\\20\neq 68$

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: $(x+2)^2+y^2=20$

• R(0,4):

$(x+2)^2+y^2=20\\(0+2)^2+4^2=20\\2^2+4^2=20\\4+16=20\\20=20$

The point R(0,4) verifies the equation.

• S(6,2):

$(x+2)^2+y^2=20\\(6+2)^2+2^2=20\\8^2+4^2=20\\64+16=20\\80\neq 20$

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: $(x-2)^2+y^2=20$

• R(0,4):

$(x-2)^2+y^2=20\\(0-2)^2+4^2=20\\(-2)^2+16=20\\4+16=20\\20=20$

The point R verifies the equation.

• S(6,2):

$(x-2)^2+y^2=20\\(6-2)^2+2^2=20\\4^2+4=20\\16+4=20\\20=20$

The point S verifies the equation.

• T(4,-4):

$(x-2)^2+y^2=20\\(4-2)^2+(-4)^2=20\\2^2+16=20\\4+16=20\\20=20$

The point T verifies the equation.

• U(-2,-2):

$(x-2)^2+y^2=20\\(-2-2)^2+(-2)^2=20\\(-4)^2+4=20\\16+4=20\\20=20$

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: $(x-2)^2+y^2=20$

We can see that the graph of the equation verifies the answer.