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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. |
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Answer» Given circles are S1 = x2 + y2 + 4x + 2y + 1 = 0 S2 = x2 + y2 – 2x + 6y – 6 = 0 S1 –S2 = 0 6x – 4y + 7 = 0 System of co-axial circle is S1 + λ(S1 –S2) = 0 x2 + y2 + 4x + 2y + 1 + λ(6x – 4y + 7) = 0 x2 + y2 + 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(x2 + y2) + 98x + 56y + 19 = 0 |
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