Question

Find the coordinates of the centres, and the radii, of the four circles which touch the sides of the triangle the coordinates of whose angular points are the points $(6,0)$, $(0,6)$, and $(7,7)$.

Answer 1

$BC=a=\sqrt{{(6-0)}^2+{(0-6)}^2}=6\sqrt{2}AC=b=\sqrt{{(7-6)}^2+{(7-1)}^2}=5\sqrt{2}AB=c=\sqrt{{(7-0)}^2+{(7-6)}^2}=5\sqrt{2}$

Coordinates of incentre are

$\left(\frac{aA+bB+cC}{a+b+c}\right)(\frac{7\times 6\sqrt{2}+6\times 5\sqrt{2}+0\times 5\sqrt{2}}{6\sqrt{2}+5\sqrt{2}+5\sqrt{2}},\frac{7\times 6\sqrt{2}+0\times 5\sqrt{2}+6\times 5\sqrt{2}}{6\sqrt{2}+5\sqrt{2}+5\sqrt{2}})I(\frac{9}{2},\frac{9}{2})$

Equation of $BC$ is

$y-0=\frac{6-0}{0-6}(x-6)x+y=6$

Inradius $=$ perpendicular distance of $I$ from $BC$

$r=\frac{\frac{9}{2}+\frac{9}{2}-6}{\sqrt{1^2+1^2}}=\frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}$

Coordinates of $O_1$ are

$\left(\frac{-aA+bB+cC}{-a+b+c}\right)(\frac{-7\times 6\sqrt{2}+6\times 5\sqrt{2}+0\times 5\sqrt{2}}{-6\sqrt{2}+5\sqrt{2}+5\sqrt{2}},\frac{-7\times 6\sqrt{2}+0\times 5\sqrt{2}+6\times 5\sqrt{2}}{-6\sqrt{2}+5\sqrt{2}+5\sqrt{2}})O_1(-3,-3)$

radius $=$ perpendicular distance of $O_1$ from $BC$

$r_1=\frac{|-3-3-6|}{\sqrt{1^2+1^2}}=\frac{12}{\sqrt{2}}=6\sqrt{2}$

Coordinates of $O_2$ are

$\left(\frac{aA-bB+cC}{a-b+c}\right)(\frac{7\times 6\sqrt{2}-6\times 5\sqrt{2}+0\times 5\sqrt{2}}{6\sqrt{2}-5\sqrt{2}+5\sqrt{2}},\frac{7\times 6\sqrt{2}-0\times 5\sqrt{2}+6\times 5\sqrt{2}}{6\sqrt{2}-5\sqrt{2}+5\sqrt{2}})(2,12)$

Equation of $AC$ is

$y-6=\frac{7-6}{7-0}(x-0)x-7y+42=0$

$r_2=$perpendicular distance of $O_2$ from $AC$

$r_2=\frac{|2-12\left(7\right)+42|}{\sqrt{1^2+7^2}}=\frac{40}{5\sqrt{2}}=4\sqrt{2}$

Coordinates of $O_3$ are

$\left(\frac{aA+bB-cC}{a+b-c}\right)(\frac{7\times 6\sqrt{2}+6\times 5\sqrt{2}-0\times 5\sqrt{2}}{6\sqrt{2}+5\sqrt{2}-5\sqrt{2}},\frac{7\times 6\sqrt{2}+0\times 5\sqrt{2}-6\times 5\sqrt{2}}{6\sqrt{2}+5\sqrt{2}-5\sqrt{2}})(12,2)$

Equation of $AB$ is

$y-0=\frac{7-0}{7-6}(x-6)$

$y=7x-42$

$7x-y-42=0$

$r_3=$perpendicular distance of $O_3$ from $AB$

$r_3=\frac{|12\left(7\right)-2-42|}{\sqrt{49+1}}=\frac{40}{5\sqrt{2}}=4\sqrt{2}$