Two circles (x + a)^2 + (y + b)^2 = a^2 and ( | 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$$(x + \alpha)^2 + (y + \beta)^2 = \be

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$2(a\alpha + b\beta) = b^2 + \alpha^2$

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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$$(x + \alpha)^2 + (y + \beta)^2 = \beta^2 \implies x^2 + y^2 + 2\alpha x + 2\beta y + \alpha^2 = 0$Comparing these with the general form $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$,we get:$g_1 = a, f_1 = b, c_1 = b^2$$g_2 = \alpha, f_2 = \beta, c_2 = \alpha^2$Two circles cut orthogonally if $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Substituting the values:$2(a\alpha) + 2(b\beta) = b^2 + \alpha^2$$2(a\alpha + b\beta) = b^2 + \alpha^2$.

Two circles $(x + a)^2 + (y + b)^2 = a^2$ and $(x + \alpha)^2 + (y + \beta)^2 = \beta^2$ cut orthogonally if:

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