help here plz (1 Viewer)

[onebytwo]

a queation that a friend asked me with which i had no idea:

find the equations of the four circles that are tangent to the line x+y-2=0, the x-axis and the y-axis.

thanks

 

[onebytwo]

do the perpendicular bisectors of the sides of a triangle necessarily concurrent, or is it the bisectors of the angles that are concurrent?

 

[SoulSearcher]

Perpendicular bisectors of the sides of a triangles are concurrent.

 

[Riviet]

Just to get you started if you haven't already. Draw a number plane and sketch x+y-2=0. Now draw 4 circles which are tangential to all three of the set conditions, ie the two axes and the linear function. Start with the circle inside the triangle formed by the two axes and the linear function.

 

[onebytwo]

thats what ive done, but i just want to know, is the centroid of the triangle also the centre of the circle inscribed within that triangle

 

[SoulSearcher]

Yes.

 

[LoneShadow]

From the digram [see attachment] you can see that [using Circle geometry, specifically properties of tangents to a circle]:
(1): x1=y1 --> Let A(x1,x1)
(2): x2=y2 --> Let B(x2,x2)
(3): x3=-y3 --> Let C(x3,-x3)
(4): x4=-y4 --> Let D(x4,-x4)

Observe that the radius of each circle is the distance between it's centre and the point where it touches the line x+y-2=0. So, using the formula for distance between a point and a line:

Line: ax+by+c=0 and point(xi,yi)
Distance = Absolute value of (axi+byi+c)/sqrt(a^2+b^2)

Now, using coordinates of points A,B,C,D and line x+y-2=0, we get:
(1): r1 = sqrt(2)*(x1-1)
(2): r2 = sqrt(2)*(x2-1)
(3): r3 = sqrt(2)
(4): r4 = sqrt(2)

[Note that circles centred at C and D are the same except they have been moved in the plane]

I'll just find the equation of circle centred at A and leave the rest for you to find out. It should not be hard when you see the method:

The general formula for this circle is (x-x1)^2 + (y-y1)^2 = r1^2, which is the same as (x-x1)^2+(y-x1)^2=r1^2 [since x1=y1].
Now, when x=x1, y = 0. Substitude this in the previous formula:

(x1-x1)^2+(0-x1)^2=r1^2 => x1 = r1 = sqrt(2)*(x1-1). So x1 = 2+sqrt(2). Therefor y1 = x1 = 2+sqrt(2). So A(2+sqrt(2), 2+sqrt(2)):
Circle centred at A: (x-2-sqrt(2))^2 + (y-2-sqrt(2))^2 = 2*(1-sqrt(2))^2.

Work in the same way for the other 3 circles.