If the circles (x+a)^2+(y+b)^2=a^2 and (x+c)^ | vedclass

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(B) The given equations of the circles are:$(x+a)^2+(y+b)^2 = a^2 \implies x^2+y^2+2ax+2by+b^2 = 0 \quad \dots (i)$$(x+c)^2+(y+d)^2 = d^2 \implies x^2

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$c(2a-c)$

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(B) The given equations of the circles are:$(x+a)^2+(y+b)^2 = a^2 \implies x^2+y^2+2ax+2by+b^2 = 0 \quad \dots (i)$$(x+c)^2+(y+d)^2 = d^2 \implies x^2+y^2+2cx+2dy+c^2 = 0 \quad \dots (ii)$Comparing with the general form $x^2+y^2+2gx+2fy+c=0$:For circle $(i)$,$g_1=a, f_1=b, c_1=b^2$.For circle $(ii)$,$g_2=c, f_2=d, c_2=c^2$.Since the circles cut orthogonally,the condition is $2(g_1g_2 + f_1f_2) = c_1 + c_2$.Substituting the values:$2(ac + bd) = b^2 + c^2$$2ac + 2bd = b^2 + c^2$$2ac - c^2 = b^2 - 2bd$$c(2a - c) = b(b - 2d)$Thus,$b(b-2d) = c(2a-c)$.

If the circles $(x+a)^2+(y+b)^2=a^2$ and $(x+c)^2+(y+d)^2=d^2$ cut orthogonally,then $b(b-2d) =$

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