1.

Find the equation of the system of circles co-axial with the circles x2 + y2 + 4x + 2y + 1 = 0 and x2 + y2 –  2x + 6y – 6 = 0. Also find the equation of that particular circles whose centre lies on radical axis. 

Answer»

Given circles are

S1 = x+ y+ 4x + 2y + 1 = 0

S2 = x+ y– 2x + 6y – 6  = 0

S1 –S2 = 0

6x – 4y + 7 = 0

System of co-axial circle is S1λ(S1 –S2) = 0

x+ y+ 4x + 2y + 1 + λ(6x – 4y + 7) = 0

x+ y+ 2x(2 + 3λ) + 2y(1 – 2λ) + 1 + 7λ = 0

Centre of this circle is (–(2 + 3λ), – (1–2λ)

lies on radical axis

6( – 2 – 3λ) + 4(1 – 2λ) + 7 = 0 

– 12 – 18λ + 4 – 8λ + 7

=  –1 –26λ = 0

λ = -1/26

∴ Required particular member of co-axial circle is 26(x+ y2) + 98x + 56y + 19 = 0



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