Question

From a point on the circle x² + y2 + 2gx + 2y + c = 0 two tangents are drawn to the
circle x + y + 2gx + 2fy + c sin^2 a + (g^2 + f^2) cos^2 a = 0, prove that the angle between
them is 2a.​

Answer 1

Answer:

Proof.. to be proved.  

Find the equation of chord of contact on to the 2nd circle. Find the equation for the pair of tangents next. Then use the formula for the angle between two lines, given the equation of the pair of two straight lines.

Step-by-step explanation:

Given a circle C1 (x,y): x² + y²+ 2g x + 2f y + c = 0 ----- (1)

** (variable f was missing in the term 2fy in the given Qn)

Let P(a, b) be a point on the circle.  

So a² + b² + 2ga + 2fb + c = 0 ----- (2)

=> (a+g)² + (b+f)² = g²+f²-c   ----(3)

Given circle C2 (x,y):  

x² + y² + 2 g x + 2 f y + c Sin² A + (g²+f²) cos² A = 0.

=> x² + y² + 2 g x + 2 f y + c + (g²+f²-c) cos² A = 0. --- (4)

Here A is an angle.

The equation of the chord of contact CT(x,y), line joining the points of contact of the two tangents T1, & T2 drawn from P(a,b) on to the circle C2, is given by :

CT: a x+ b y+ g(x+a)+ f(y+b)+ c sin²A +(g²+f²) cos²A = 0 --- (5)

The pair of tangents TT(x,y) from P(a,b) onto C2(x, y) is given by the equation:

C2(a, b) * C2(x,y) = [ CT(x,y) ]²     ----- (6)

C2(a,b) = a² + b² + 2ga + 2fb + c + (g²+f²-c) Cos²A

            = (g²+f²-c) Cos²A , using (3)

The equation of pair of tangents desired is:

[(g²+f²-c)Cos²A] * [x²+y²+2gx+2fy+c+(g² +f² -c)cos²A]  

= [ax + by + g(x+a) + f(y+b) + (g²+f²-c) cos² a ]²   ---- (7)

We find the angle between two tangents without expanding all the terms. We need only three coefficients for using the formula.

 

In equation (7) the coefficient of x² term is =  

   α = (g²+f²-c) Cos²A - (a+g)²  

   α = (b+f)²-(g²+f²-c) Sin²A, by using (3)       ------ (8)

Similarly, the coefficient of y² in (7) is :

   β = (a+g)²-(g²+f²-c) Sin²A.       ------ (9)

The coefficient of xy term in (7) is :

  δ = -(a+g)(b+f)                      ---- (10)

Let us find δ² - α * β by using (8) , (9) and (10) and simplifying.

δ² - α * β = (g²+f²-c) Sin²A [-((g²+f²-c) Sin²A + (a+g)² + (b+f)²]

       = (g²+f²-c) Sin²A * (g²+f²-c) Cos²A       , by using (3)

       = (g²+f²-c)² Sin²2A * 1 /4    ----- (11)

Find α + β = (a+g)² + (b+f)² - 2 (g²+f²-c) Sin² A.

So α + β = (g²+f²-c) [ 1 - 2 Sin² A] , using (3).

      = (g²+f²-c) Cos 2A   --- (12)

The angle θ between the two lines in the pair of tangents is given by:

Tan θ = [2 √(δ² - α*β)] / (α + β)              ---- (13)

      = Sin 2A / Cos 2A   by simplifying  

      = Tan 2A

Hence, the angle = θ = 2A.  

*****  ANSWER .

Formula used:

Generic equation for a pair of straight lines :

   a x² + b y²+ 2h xy + 2g x + 2f y + c = 0.   --- (14)

Angle between the pair of lines:  

  Tan θ =  [2 √(h² -ab)] / (a+ b)       --- (15)