How do you show that two circles #x^2+y^2-2hx-2ky+c = 0# and #x'^2+y'^2-2h'x-2k'y+c'=0# intersect at right angles if and only if #2hh'+2kk' = c+c'# ?

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Answer 1 · 2015-09-19T08:56:58

Find the condition for a triangle formed by the centres of the circles and a point of intersection to be a right angled triangle.

#x^2+y^2-2hx-2ky+c#

#= (x-h)^2+(y-k)^2 + (c-h^2-k^2)#

So this is the equation of a circle with centre #(h, k)# and radius:

#sqrt(h^2+k^2-c)#

The distance between the centres of the two circles is:

#sqrt((h-h')^2 + (k-k')^2)#

So the triangle formed by the centres of the two circles and a point of intersection has sides of length:

#sqrt(h^2+k^2-c)#, #sqrt(h'^2+k'^2-c')#

and #sqrt((h-h')^2 + (k-k')^2)#

This will be a right angled triangle if and only if it satisfies Pythagoras:

#(h^2+k^2-c) + (h'^2+k'^2-c')#

#=(h-h')^2 + (k-k')^2#

#= h^2 - 2hh' + h'^2 + k^2 - 2kk' + k'^2#

Subtract #h^2+k^2+h'^2+k'^2# from both ends to find:

#-c-c' = -2hh'-2kk'#

Multiply both sides by #-1# to get:

#2hh'+2kk' = c+c'#