Consider circle S' = 0, whose centre lies on line 2y = 2x + 11 and cuts circle S : x2 + y2 = 9 orthogonally, then circle S' = 0 always passes through fix point

Engineering

Mathematics

Family of Circle

Question

Consider circle S' = 0, whose centre lies on line 2y = 2x + 11 and cuts circle S : x² + y² = 9 orthogonally, then circle S' = 0 always passes through fix point

$(\frac{- 9}{2}, \frac{- 9}{2})$

(1, –1)

(–1, 1)

$(\frac{9}{2}, \frac{- 9}{2})$

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S' = x² + y² + 2gx + 2fy + c = 0,

where – 2f = – 2g + 11 ⇒ 2f = 2g – 11 and c = 9

$∴$ S' = x² + y² + 2gx + (2g – 11) y + 9 = 0

$∴$ S' = (x² + y² – 11y + 9) + g (2x + 2y) = 0

$∴$ Fix point is (– 1, 1) and $(\frac{- 9}{2}, \frac{9}{2})$ .