Two given circles x^2 + y^2 + ax + by + c = 0 | vedclass

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(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Comparing the given circles with the general form: For the first circle,$2g_1

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$ad + be = 2(c + f)$

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(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Comparing the given circles with the general form: For the first circle,$2g_1 = a$,$2f_1 = b$,and $c_1 = c$. For the second circle,$2g_2 = d$,$2f_2 = e$,and $c_2 = f$. The condition for two circles to intersect orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. Substituting the values: $2(\frac{a}{2})(\frac{d}{2}) + 2(\frac{b}{2})(\frac{e}{2}) = c + f$ $\frac{ad}{2} + \frac{be}{2} = c + f$ Multiplying by $2$ on both sides: $ad + be = 2(c + f)$.

Two given circles $x^2 + y^2 + ax + by + c = 0$ and $x^2 + y^2 + dx + ey + f = 0$ will intersect each other orthogonally,only when

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