Question
Tangents are drawn to the circle $x^{2} + y^{2} = 1$ at the points where it is met by the circles, $x^{2} + y^{2} - (\lambda + 6)x + (8 - 2\lambda)y - 3 = 0.\lambda$ being the variable. The locus of the point of intersection of these tangents is:
Options
Solution
[Key Concept: Tangents at the points where a line meets the circle: use the chord of contact / pole of the line.]
Equation of $AB$ is
$$\begin{array}{r} xh + yk = 1\#(1) \\ AB \equiv S_{1} - S_{2} = 0\#(1) \\ \Rightarrow \ (\lambda + 6)x + (2\lambda - 8)y + 2 = 0\#(2) \end{array}$$
(1) & (2) are identical
$${\Rightarrow \ \frac{h}{\lambda + 6} = \frac{k}{2\lambda - 8} = \frac{- 1}{2} }{\Rightarrow \frac{2h}{2\lambda + 12} = \frac{k}{2\lambda - 8} = \frac{2h - k}{20}}$$
$$= - \frac{1}{2}\left( \text{~}\text{as}\text{~}\frac{a}{b} = \frac{c}{b} = \frac{a - c}{b - d} \right)$$
$$\begin{matrix} & \ \Rightarrow \ 2x - y = - 10 \\ & \ \Rightarrow \ 2x - y + 10 = 0 \end{matrix}$$
Create a free account to view solution
View Solution Free