A circle passes through two adjacent vertices of a square and is tangent to one side of the

square. If the side length of the square is 2, what is the radius of the circle?

need answer and explination as soon as possible please!!! Thank you so much!!!

Trinityvamp286 Jul 15, 2017

#1**+1 **

This is ** a** way to draw a circle that passes through two adjacent vertices of a square and is tangent to one side of the square....I think......

radius = diameter / 2

radius = side length of square / 2

radius = 2 / 2

radius = 1

hectictar Jul 15, 2017

#2**+2 **

As long as the points aren't collinear, we can construct a circle that will pass through any three points......see the pic below.......

Call a, b the center of the circle

And let the circle pass through points (0,2), (2,2) and touch point (1,0)

And we need to equate the distances from (a,b) to (1,0) and from (a,b) to (2,2)

(a-1)^2 + (b)^2 = (a-2)^2 + (b - 2)^2

Expanding this we have

a^2 - 2a + 1 + b^2 = a^2 - 4a + 4 + b^2 - 4b + 4 simplify

-2a + 1 = -4a + 4 - 4b + 4

2a + 4b = 7 (1)

And then we can equate the distances from (a,b) to (1,0) and from (a,b) to (0,2)

(a-1)^2 + (b)^2 = (a)^2 + (b - 2)^2 simplify

a^2 - 2a + 1 + b^2 = a^2 + b^2 - 4b + 4

-2a + 4b = 3 (2)

Adding (1) and (2) we have that

8b =10 → b = 5/4

Subbing this back into (1)

2a + 4(5/4) = 7

2a + 5 = 7 → a = 1

So the center of the required circle is ( 1, 5/4)

And the distance that this point is from (1, 0) is just 5/4 - (1.25) - which is the radius

Here's a pic :

(a,b), represented by E, is the center........The circle passes through B, C and is tangent to AD

CPhill Jul 15, 2017