equation of a cirle?

Question

A triangle has its sides on the lines x+2y-5=0, 2x +y+2=0&2x-y-10=0. Find equation of the circle inscribe the triangle?

Answer 1

Rewriting these we get:

(a) y=(5-x)/2 (b) y=-2x-2 (c) y=2x-10 for lines a, b and c.

Lines a and c are perpendicular because the slope of a is -½ while the slope of c is 2, which is the negative inverse of -½.

Lines a and b intersect when (5-x)/2=-2x-2, 5-x=-4x-4, 3x=-9, x=-3 so y=(5+3)/2=6-2=4, the point A(-3,4). The slope of b is -2.

Lines a and c intersect when (5-x)/2=2x-10, 5-x=4x-20, 5x=25, x=5 so y=0, the point B(5,0). Angle B is a right angle.

Lines b and c intersect when -2x-2=2x-10, 4x=8, x=2 so y=-6, the point C(2,-6).

Since the slopes of b and c have the same magnitude but slope in the opposite direction to one another, their bisector (of angle C) is vertical, implying that x-coordinate is the same as that of point C, x=2.

Slope of b is -2 and the slope of a is -½. These correspond to two angles θ=arctan(-2) and φ=arctan(-½). The slope of the bisector is the average of these two slopes=½(θ+φ)=½(arctan(-2)+arctan(-½)).

tan(θ+φ)=(tanθ+tanφ)/(1-tanθtanφ)=(-2-½)/(1-(-2)(-½)). Note that these angles are negative because their tangents are negative. A negative angle simply means we take the supplementary angle (obtuse). But 1-(-2)(-½)=0 so θ+φ=-90°, and ½(θ+φ)=-45° or 135°, the slope of the bisector of angle A=tan(-45)=-1. We have the slope of the bisector and we know it passes through A so we can find its equation:

y-4=-1(x+3)=-x-3, y=-x+1.

We also know that the centre of the inscribed circle is the point of intersection of the angle bisectors and we know the x-coordinate of the bisector of angle C is 2, so the y-coordinate is y=-2+1=-1. The centre of the incircle=O(2,-1). To find the radius we need the length of the perpendicular from O to AB. This perpendicular has to be parallel to BC (segment of 2x-y-10=0, line c) because angle B is a right angle. The perpendicular has a slope of 2 and passes through O so we can find its equation:

y+1=2(x-2), y+1=2x-4, y=2x-5, and it meets BC when 2x-5=(5-x)/2, 4x-10=5-x, 5x=15, x=3 so y=1, the point P(3,1).

The radius OP=√((3-2)²+(1-(-1))²)=√(1+4)=√5, so the equation of the incircle is:

(x-2)²+(y+1)²=5.