# A(4,4) and B(-4,4) are two points on the circle x^2+y^2=32 and P is any point on the circle.Prove that the angular bisector of angle APB pass through a fixed point.

148 Points
14 years ago

Dear sudarshan

angular bisector of angel must made 45/2 angel with each line AP and BP

we know

tan 45 = 2tan45/2 /(1- tan245/2)

let tan 45/2  = m

so 1= 2m/(1-m2)

m2 +2m -1 =0 ................1

now slope of line AP (m1) = (4- 4√2sinθ)/((4- 4√2cosθ)

slope of bisector m2 = (y- 4√2sinθ)/((x- 4√2cosθ)

so tan45/2 =m= (m1-m2)/(1+m1m2)

put value of m1 and m2 and simplify

√2 cosθ ( -mx-4m +y -4) + √2sinθ (-my -4m+4-x) +(mx +my +32m -y +x) =0

solve coefficiet of cosθ and sinθ  and find value of x and y and put it in constant term if it satiesfied it then this line always pass throug this point for all value of θ.

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.

All the best.

Regards,