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(c) Given $\frac{{{T_1}}}{{{T_2}}} = \frac{4}{3}$, where $T_1$ and $T_2$ are the length of tangents drawn to the given circle. $ \Rightarrow \fra

Vedclass · Accepted Answer

$-21/ 4$

Vedclass · Answer

(c) Given $\frac{{{T_1}}}{{{T_2}}} = \frac{4}{3}$, where $T_1$ and $T_2$ are the length of tangents drawn to the given circle. $ \Rightarrow \frac{{\sqrt {1 + 4 + 1 + 2 - 4} }}{{\sqrt {{{(1)}^2} + {{(2)}^2} - \frac{1}{3} - \frac{2}{3} + \frac{k}{3}} }} = \frac{4}{3} \Rightarrow k = - \frac{{21}}{4}$.

If the length of the tangents drawn from the point $(1,2)$ to the circles ${x^2} + {y^2} + x + y - 4 = 0$ and $3{x^2} + 3{y^2} - x - y + k = 0$ be in the ratio $4 : 3$, then $k =$

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