The equation of a circle which cuts the three circles

MATHEMATICS · Single correct · CBSE · Mathematics · Conic Sections · Conic Sections

Problem

The equation of a circle which cuts the three circles x2+y2−3x−6y+14=0x^2 + y^2 - 3x - 6y + 14 = 0 , x2+y2−x−4y+8=0x^2 + y^2 - x - 4y + 8 = 0 and x2+y2+2x−6y+9=0x^2 + y^2 + 2x - 6y + 9 = 0 orthogonally is (1) x2+y2−2x−4y+1=0x^2 + y^2 - 2x - 4y + 1 = 0 (2) x2+y2+2x+4y+1=0x^2 + y^2 + 2x + 4y + 1 = 0 (3) x2+y2−2x+4y+1=0x^2 + y^2 - 2x + 4y + 1 = 0 (4) x2+y2−2x−4y−1=0x^2 + y^2 - 2x - 4y - 1 = 0

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