Answers

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

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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)

2 4 2
click on thanks button above please
select best answer please
this was not easy the solution... did you get it ? is that right ?