Question

A circle is inscribed in a square whose vertices have coordinates R(0, 4), S(6, 2), T(4, -4), and U(-2, -2). Find the equation of the circle.

Answer 1

The center of the circle is the center of the square which cab be found by determining the midpoint of one of the diagonal, and the radius will be the half of the length of the circle.

Answer 2

what is the mid point formula?

Answer 3

First, give us one of the diagonals

Answer 4

what do you mean ? Im not exactly sure

Answer 5

You have a square , when you put the vertices in the plan you can recognize the two diagonals of the square . So the center of the circle is the midpoint of one ot that two diagonals ! Is is clear to you ?

Answer 6

not really.. i plotted it out, if i drew two diagonals across my square where they intersect would be my midpoint?

Answer 7

A.(x - 2)² + y² = 68
B.(x + 2)² + y² = 20
C.(x - 2)² + y² = 10

Answer 8

I drew the square, and I found that RT and US are the diagonals of the square.
So, if C(x,y) is the midpoint then
\[x=\frac{x_R+x_T}{2}=\frac{0+4}{2}=2\\y=\frac{y_R+y_T}{2}=\frac{4-4}{2}=0\]
So the centre is
\[C(2,0)\]
Now we calculate the length a of the square :
\[a=RU=\sqrt{(x_U-x_R)^2+(y_U-y_T)^2}=\sqrt{(-2-0)^2+(-2-4)^2}=\sqrt{40}\]
hence :
\[r=\frac a2=\frac{\sqrt{40}}{2}\]
So the equation of the circle is :
\[(x-x_C)^2+(y-y_C)^2=r^2\]
that is :
\[(x-2)^2+y^2=\frac{40}4\]
or, in another way :
\[(x-2)^2+y^2=10\]