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10-1.Circle and System of Circles
hard
બિંદુ $(a, b)$ માંથી પસાર થતા તથા વર્તૂળ ${x^2} + {y^2} = {p^2}$ ને લંબચ્છેદી હોય તેવા વર્તૂળના કેન્દ્રનો બિંદુગણનું સમીકરણ મેળવો.
A
$2ax + 2by - ({a^2} + {b^2} + {p^2}) = 0$
B
$2ax + 2by - ({a^2} - {b^2} + {p^2}) = 0$
C
${x^2} + {y^2} - 3ax - 4by + ({a^2} + {b^2} - {p^2}) = 0$
D
${x^2} + {y^2} - 2ax - 3by + ({a^2} - {b^2} - {p^2}) = 0$
(AIEEE-2005) (IIT-1988)
Solution
(a) Let equation of circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$ with ${x^2} + {y^2} = {p^2}$ cutting orthogonally,
we get $0 + 0 = + c – {p^2}$
or $c = {p^2}$ and passes through $(a, b)$,
we get ${a^2} + {b^2} + 2ga + 2fb + {p^2} = 0$ or
$2ax + 2by – ({a^2} + {b^2} + {p^2}) = 0$
Required locus as centre $( – g,\; – f)$ is changed to $(x, y)$.
Standard 11
Mathematics
