Tangents are drawn to the circle x^2 + y^2 = | vedclass

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Question

(A) Let the point of intersection of the tangents be $(x_1, y_1)$. The chord of contact of the tangents from $(x_1, y_1)$ to the circle $x^2 + y^2 = 1

Vedclass · Accepted Answer

$2x - y + 10 = 0$

Vedclass · Answer

(A) Let the point of intersection of the tangents be $(x_1, y_1)$. The chord of contact of the tangents from $(x_1, y_1)$ to the circle $x^2 + y^2 = 1$ is given by $xx_1 + yy_1 = 1$,or $xx_1 + yy_1 - 1 = 0$ $... (1)$.The common chord of the two circles $S_1: x^2 + y^2 - 1 = 0$ and $S_2: x^2 + y^2 - (\lambda + 6)x + (8 - 2\lambda)y - 3 = 0$ is given by $S_1 - S_2 = 0$.$(x^2 + y^2 - 1) - (x^2 + y^2 - (\lambda + 6)x + (8 - 2\lambda)y - 3) = 0$$(\lambda + 6)x - (8 - 2\lambda)y + 2 = 0$ $... (2)$.Since the chord of contact is the same as the common chord,we compare equations $(1)$ and $(2)$:$\frac{x_1}{\lambda + 6} = \frac{y_1}{-(8 - 2\lambda)} = \frac{-1}{2}$From $\frac{x_1}{\lambda + 6} = -\frac{1}{2}$,we get $\lambda + 6 = -2x_1 \Rightarrow \lambda = -2x_1 - 6$.From $\frac{y_1}{2\lambda - 8} = -\frac{1}{2}$,we get $2\lambda - 8 = -2y_1 \Rightarrow \lambda = -y_1 + 4$.Equating the two expressions for $\lambda$:$-2x_1 - 6 = -y_1 + 4$$y_1 - 2x_1 = 10$$2x_1 - y_1 + 10 = 0$.Replacing $(x_1, y_1)$ with $(x, y)$,the locus is $2x - y + 10 = 0$.

Tangents are drawn to the circle $x^2 + y^2 = 1$ at the points where it is met by the circles $x^2 + y^2 - (\lambda + 6)x + (8 - 2\lambda)y - 3 = 0$,where $\lambda$ is a variable. The locus of the point of intersection of these tangents is:

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