19.0 Orthogonal Circles

Assume two intersecting circles with equation of circles are $${S_1} \equiv {x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$$ and $${S_2} \equiv {x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$$

When two circles intersect at right angle i.e., $\angle {C_1}P{C_2} = {90^ \circ }$, then they are called orthogonal circles and they satisfy the condition $${c_1} + {c_2} = 2{g_1}{g_2} + 2{f_1}{f_2}$$

Proof: From figure $53$, using Pythagoras theorem in $\Delta {C_1}{C_2}P$, ${r_1}^2 + {r_2}^2 = {\left( {{C_1}{C_2}} \right)^2}$ and use distance formulae, to find distance between ${C_1}$ and ${C_2}$, we get $$({\sqrt {{g_1}^2 + {f_1}^2 - {c_1})} ^2} + ({\sqrt {{g_2}^2 + {f_2}^2 - {c_2})} ^2} = {g_1}^2 + {g_2}^2 - 2{g_1}{g_2} + {f_1}^2 + {f_2}^2 - 2{f_1}{f_2}$$ $${c_1} + {c_2} = 2{g_1}{g_2} + 2{f_1}{f_2}$$

NOTE: Equation of a circle cutting the three circles ${x^2} + {y^2} + 2{g_i}x + 2{f_i}y + {c_i} = 0{\text{ (i = 1,2,3)}}$ orthogonally is $$\left| {\begin{array}{c}{{x^2} + {y^2}}&x&y&1 \\{ - {c_1}}&{{g_1}}&{{f_1}}&{ - 1} \\{ - {c_2}}&{{g_2}}&{{f_2}}&{ - 1} \\{ - {c_3}}&{{g_3}}&{{f_3}}&{ - 1}\end{array}} \right|$$