If the circles x^2+y^2-2 lambda x-2 y-7=0 and | vedclass

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(D) The equations of the circles are:$x^2+y^2-2 \lambda x-2 y-7=0$  $(i)$$3(x^2+y^2)-8 x+29 y=0 \Rightarrow x^2+y^2-\frac{8}{3} x+\frac{29}{3} y=0$  $

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(D) The equations of the circles are:$x^2+y^2-2 \lambda x-2 y-7=0$  $(i)$$3(x^2+y^2)-8 x+29 y=0 \Rightarrow x^2+y^2-\frac{8}{3} x+\frac{29}{3} y=0$  $(ii)$Comparing these with the general form $x^2+y^2+2gx+2fy+c=0$:For circle $(i)$: $g_1 = -\lambda, f_1 = -1, c_1 = -7$For circle $(ii)$: $g_2 = -\frac{4}{3}, f_2 = \frac{29}{6}, c_2 = 0$Since the circles are orthogonal,the condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Substituting the values:$2(-\lambda)(-\frac{4}{3}) + 2(-1)(\frac{29}{6}) = -7 + 0$$\frac{8}{3}\lambda - \frac{29}{3} = -7$$\frac{8}{3}\lambda = -7 + \frac{29}{3} = \frac{-21+29}{3} = \frac{8}{3}$$\lambda = 1$

If the circles $x^2+y^2-2 \lambda x-2 y-7=0$ and $3(x^2+y^2)-8 x+29 y=0$ are orthogonal,then $\lambda$ is equal to

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