System of co-axial circles whose limiting points are given
Let (α,β ) and ( γ ,δ ) be the two given limiting points
Then corresponding circles with zero radii are
(x–α)2 +(y–β )2 = 0 = x2 +y2 –2αx–2β y+α2 +β2 = 0
(x– γ )2 +(y–δ )2 = 0 = x2 +y2 –2 γ x–2 δ y+ γ2 +δ2 = 0
System of co-axial circle equation is
x2 +y2 –2αx–2β y+α2 +β2 +λ(x2 +y2 –2 γ x+2 δ y+ γ2 +δ2 ) = 0 (λ ≠ –1)
centre of this circle is
\(\left(\frac{\alpha+γλ}{1+λ},\frac{β +δλ}{1+λ
}\right)------(1)\)
and radius= \(\sqrt{\left(\frac{α +γλ}{1+λ}\right)^2+\left(\frac{β+δλ}{1+λ}\right)^2-\frac{(α^2 + β^2 +λγ^2+λδ^2)}{1+λ}}=0\)
After solving find λ substitute in (1)
We get the limiting point of co-axial system
