Question

prove tha the locus of the point of interfsection of the line drawn through the points (a,0) , (-a,0) with include a constant angle β is the circle
x^2+y^2-a^2±2acotβ

Answer 1

step1
let the moving point be (h,k)

Answer 2

step2
so the slope of the line joining (h,k) and (a,0) is (k-0)/(h-a)
so the slope of the line joining (h,k) and (-a,0) is (k-0)/(h+a)

Answer 3

step 3
By the given condition we have
+- tanβ = (k/(h-a) - k/(h+a) )/(1 + k^2/(h^2-a^2)..

Answer 4

step 4
simplify the equation both sides and replace h by x and k by y to get the desired equation

Answer 5

hence we have
+- tanβ = 2ak /(h^2+k^2-a^2)
or
(h^2+k^2-a^2) +- 2akcotβ =0
or
x^2+y^2 -a^2+-2aycotβ =0

Answer 6

thank you very much but could you explain step 3 please?

Answer 7

yw

Answer 8

in step3
the angle say β between the lines having slopes m1 and m2 is given by
+- tanβ = (m1-m2)/(1+m1*m2)

Answer 9

ok thank you

Answer 10

yw