# The locus of poles of tangents to the circle x^2+y^2=a^2 with respect to the circle x^2+y^2+2ax-a^2=0 is

1
by 2092000

2015-09-18T17:39:35+05:30

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Circle1:  x² + y² = a²
let A (x1, y1) be a point on circle 1:      so  x1² + y1² = a²      --- (1)
Tangent T  to circle at point A(x1, y1):    x * x1 + y * y1 = a²    --- (2)
slope of tangent T at  A:  - x1/ y1
Slope of Normal N to tangent T at A:   y1/x1

Circle 2:  (x +a)² + y² =  2 a²
radius R = √2 a        center O = (-a , 0)

Equation of straight line OQ  parallel to Normal N and passing through O (-a, 0) is:
y = (x+a) * y1/x1          -- (3)

OQ = perpendicular distance from O onto the line T
= | x1 *(-a) + y1 * 0  - a² | / √[x1²  + y1² ]              --- using (2)
= a |x1 + a| / a
=   x1 + a          as  x1 <= a

Let P (x2, y2) be the pole of  line T  wrt  circle 2.
OP²  =    (x2+a)² + y2²    -- (4)

The pole  P of tangent T to circle 1,  wrt Circle 2, will lie on the Normal N and at a distance OP from center O :
OP = R² / OQ      = 2 a² / (x1 + a)      --- (5)

As  P(x2, y2) lies on OQ,  by (3) we get,
y2 = (x2 + a) * y1 / x1          -- (6)

From  (4) and (5) we get:
(x2 + a)² + y2² = 4 a⁴ / (x1 + a)²
=>  y2² [ a² / y1² ] =   4 a⁴ / (x1 +a)²          using    (6) and (1)
=>  y2² = 4 a² y1² / (x1 + a)²

let us use parametric representation for  circle 1:  x1 = a cos Ф      and  y1 = a sin Ф
y2² = 4 a² a² Sin² Ф / (2 a² Cos²Ф/2)²  =  a² Sin² Ф / Cos²Ф/2
= 4 a² Sin² Ф/2
So  y2 = + or -  2a Sin Ф/2          --- (7)

Cos²Ф/2 = 1 - SIn²Ф/2 = (4a² - y2²) /4a²
tan Ф/2 =  + -  y2 / √(4a² - y2²)
tan Ф =  + 2 y2 √(4 a² - y2² ) / [ 4a²  -  2 y2²]
CotФ  = (2a² - y2²) / [ y2 √ (4a² - y2²)  --- (8)

From (6) we have :
(x2 + a) = y2 * cot Ф  =  (2 a² - y2² ) / √(4a² - y2²)

Now for getting the locus of the pole P(x2, y2), replace them by (x, y):

(x + a)²  =  (2 a² - y²)² / (4 a² - y²)      --- (9)

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