Refer to figure below. (Obs.: D is the circle's center)
I created this figure using MS Excel
With A(1,5), B(9,4) and C(1,8)A(1,5),B(9,4)andC(1,8)
AB=sqrt((9-1)^2+(4-5)^2)=sqrt(64+1)=sqrt(65)~=8.1AB=√(9−1)2+(4−5)2=√64+1=√65≅8.1
BC=sqrt((1-9)^2+(8-4)^2)=sqrt(64+16)=sqrt(80)=4sqrt(5)~=8.9BC=√(1−9)2+(8−4)2=√64+16=√80=4√5≅8.9
CA=sqrt((1-1)^2+(5-8)^2)=3CA=√(1−1)2+(5−8)2=3
Using the variables m, n and om,nando
m+n=4sqrt(5)m+n=4√5
m+o=sqrt(65)m+o=√65
n+o=3n+o=3 => o=3-no=3−n
=> m+3-n=sqrt(65)m+3−n=√65 => m-n=sqrt(65)-3m−n=√65−3
Adding the last with the first equation, we get
2m=4sqrt(5)+sqrt(65)-32m=4√5+√65−3 => m=2sqrt(5)+sqrt(65)/2-1.5m=2√5+√652−1.5
In the figure we can see that
tan alpha=r/mtanα=rm => r=m*tan alphar=m⋅tanα
We only need to know alphaα
Using vectors
BvecA=(1-9)hat i+(5-4)hat j=-8hat i+hat jB→A=(1−9)ˆi+(5−4)ˆj=−8ˆi+ˆj
BvecC=(1-9)hat i+(8-4)hat j=-8hat i+4hat jB→C=(1−9)ˆi+(8−4)ˆj=−8ˆi+4ˆj
cos (2alpha)=(BvecA*BvecC)/(|BA|*|BC|)=(8*8+1*4)/(sqrt(65)*4sqrt(5))=68/(20sqrt(13))=17/(5sqrt(13))cos(2α)=B→A⋅B→C|BA|⋅|BC|=8⋅8+1⋅4√65⋅4√5=6820√13=175√13
=> 2alpha=19.440^@2α=19.440∘ => alpha=9.720^@α=9.720∘
Using the Law of Sines
3/sin(180^@-2alpha)=sqrt(65)/sin(180^@-2beta)=(4sqrt(5))/sin(180^@-2gamma)3sin(180∘−2α)=√65sin(180∘−2β)=4√5sin(180∘−2γ)
Since sines of supplementary angles are equal we can rewrite the previous equation as
3/(sin 2alpha)=sqrt(65)/(sin 2beta)=(4sqrt(5))/(sin 2gamma)3sin2α=√65sin2β=4√5sin2γ
3/(sin 2alpha)=sqrt(65)/(sin 2beta)3sin2α=√65sin2β => sin 2beta=sqrt(65)/3*sin 2alphasin2β=√653⋅sin2α
We know also that 2alpha+2beta+2gamma=180^@2α+2β+2γ=180∘ => 2gamma=180^@-2alpha-2beta2γ=180∘−2α−2β
So
3/sin(2alpha)=(4sqrt(5))/sin(180^@-2alpha-2beta)=(4sqrt(5))/sin (2alpha+2beta)=(4sqrt(5))/(sin 2alpha *cos 2beta+sin 2beta *cos 2alpha)3sin(2α)=4√5sin(180∘−2α−2β)=4√5sin(2α+2β)=4√5sin2α⋅cos2β+sin2β⋅cos2α
-> 3/cancel (sin2alpha)=(4sqrt(5))/(cancel(sin 2alpha)*sqrt(1-65/9*sin^2 2alpha)+sqrt(65)/3*cancel(sin 2alpha) *sqrt(1-sin^2 2alpha)
-> cancel(3)*sqrt(9-65sin^2 2alpha)/cancel(3)+cancel(3)*sqrt(65)/cancel(3)*sqrt(1-sin^2 2alpha)=4sqrt(5)
-> (sqrt(9-65sin^2 2alpha))^2=(4sqrt(5)-sqrt(65)*sqrt(1-sin^2 2alpha))^2
-> 9-cancel(65sin^2 2alpha)=80-8*5*sqrt(13)*sqrt(1-sin^2 2alpha)+65-cancel(65sin^2 2alpha)
40sqrt(13)*sqrt(1-sin^2 2alpha)=136
1300-1300*sin^2 2alpha=1156 => sin^2 2alpha=144/1300 => sin 2alpha=6/(5sqrt(13)) => 2alpha=19.440^@ => alpha=9.720^@
Finally,
r=m*tan alpha=(2sqrt(5)+sqrt(65)/2-1.5)*tan 9.720^@ => r=1.220