Question

if the length of the tangent from the point P to the circle x^2+y^2=a^2 is t times the length of the tangent from P to the circle x^2+y^2-2ax=0 (t isnt equal to 1), show that P lies on a fixed circle which passes through 2 common points of the intersection of the first two circles

Answer 1

I'd start by equating the two functions and finding the two common points. I'm guessing point P is some arbitrary point that doesn't have an actual numeric value.

Answer 2

1st circle has centre (0,0) and radius a

2nd circle has centre ( 1, 0) and radius 1 (found by completing the square)

Answer 3

i think you oughta start by finding the length of the tangents by the circle theorems

Answer 4

yep P is an arbitrary point.

Answer 5

the points of contact between the circles are
\[(-2a, \sqrt{3a}) (-2a, - \sqrt{3a})\]

Answer 6

oh ok....thank you....