Question

A circle S = 0 passes through points of intersection of circles $x^2 + y^2 - 2x + 4y - 1 = 0$ and $x^2 + y^2 + 4x - 2y - 5 = 0$ and cuts the circle $x^2 + y^2 = 4$ orthogonally. Then length of tangent from origin on circle S = 0 is $L$, then $\dfrac{L}{5}$ is

Answer 1

$\displaystyle \tfrac{2}{5}$

Answer 2

Solution Outline

1. Family of circles through intersections
   Let the required circle be

   $$C_1 + \lambda\,C_2 = 0$$

   where

   $$C_1: x^2 + y^2 - 2x + 4y - 1 = 0,\quad C_2: x^2 + y^2 + 4x - 2y - 5 = 0.$$

   This gives

   $$(1+\lambda)(x^2+y^2) +(-2+4\lambda)x +(4-2\lambda)y +(-1-5\lambda)=0.$$

2. Orthogonality condition
   For orthogonality with $x^2 + y^2 - 4 = 0$, the constants must satisfy

   $$c + c_1 = 0,\quad c_1 = -4.$$

   Here $c = \frac{-1 -5\lambda}{1+\lambda}$.
   So

   $$\frac{-1 -5\lambda}{1+\lambda} -4 = 0 \quad\Longrightarrow\quad -1 -5\lambda = 4(1+\lambda) \quad\Longrightarrow\quad \lambda = -\frac{5}{9}.$$

3. Equation of required circle
   Substitute $\lambda = -\tfrac{5}{9}$. Then $1+\lambda = \tfrac{4}{9}$ and

   $$2g = \frac{-2+4\lambda}{1+\lambda} = -\frac{19}{2},\quad 2f = \frac{4-2\lambda}{1+\lambda} = \frac{23}{2},\quad c = 4.$$

   So the circle is

   $$x^2+y^2 - \frac{19}{2}x + \frac{23}{2}y + 4 = 0.$$

4. Length of tangent from the origin
   For $x^2+y^2+2gx+2fy+c=0$, the tangent length from $(0,0)$ is

   $$L = \sqrt{S(0,0)} = \sqrt{c} = \sqrt{4} = 2.$$

Hence

$$\frac{L}{5} = \frac{2}{5}.$$