The value of lambda , for which the circle {x | vedclass

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Question

(d) If two circles ${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ and $2({g_1}{g_2} + {f_1}{f_2}) = {c_1} + {c_2}$ intersect orthogonally then,

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Vedclass · Accepted Answer

$\frac{{ - 5}}{4}$

Vedclass · Answer

(d) If two circles ${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ and $2({g_1}{g_2} + {f_1}{f_2}) = {c_1} + {c_2}$ intersect orthogonally then,

they must follow $2({g_1}{g_2} + {f_1}{f_2}) = {c_1} + {c_2}$ and ${g_1} = \lambda ,\,{f_1} = 3,\,{c_1} = 1$

and ${g_2} = 2,\,{f_2} = 1,\,{c_2} = 0$

So, $2\,(2\lambda + 3) = 1 + 0$

$\Rightarrow 2\lambda + 3 = \frac{1}{2}$

$ \Rightarrow \lambda = \frac{{ - 5}}{4}$.

The value of $\lambda $, for which the circle ${x^2} + {y^2} + 2\lambda x + 6y + 1 = 0$, intersects the circle ${x^2} + {y^2} + 4x + 2y = 0$ orthogonally is

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