If the point (2, lambda) lies inside the circ | vedclass

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(D) Let the given circles be $S_1 \equiv x^2+y^2-13=0$ and $S_2 \equiv x^2+y^2+x-2y-14=0$.For a point $(x_1, y_1)$ to lie inside a circle $S \equiv x^

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$(-2, 3)$

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(D) Let the given circles be $S_1 \equiv x^2+y^2-13=0$ and $S_2 \equiv x^2+y^2+x-2y-14=0$.For a point $(x_1, y_1)$ to lie inside a circle $S \equiv x^2+y^2+2gx+2fy+c=0$,we must have $S(x_1, y_1) For the first circle $S_1(2, \lambda) $2^2 + \lambda^2 - 13 $4 + \lambda^2 - 13 $\lambda^2 - 9 $(\lambda - 3)(\lambda + 3) $-3 For the second circle $S_2(2, \lambda) $2^2 + \lambda^2 + 2 - 2\lambda - 14 $4 + \lambda^2 + 2 - 2\lambda - 14 $\lambda^2 - 2\lambda - 8 $(\lambda - 4)(\lambda + 2) $-2 Taking the intersection of $(i)$ and (ii),we get:$\lambda \in (-2, 3)$.

If the point $(2, \lambda)$ lies inside the circles $x^2+y^2=13$ and $x^2+y^2+x-2y=14$,then $\lambda$ lies in the set

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