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10-1.Circle and System of Circles
hard
The centre of the circle, which cuts orthogonally each of the three circles ${x^2} + {y^2} + 2x + 17y + 4 = 0,$ ${x^2} + {y^2} + 7x + 6y + 11 = 0,$ ${x^2} + {y^2} - x + 22y + 3 = 0$ is
A
$(3, 2)$
B
$(1, 2)$
C
$(2, 3)$
D
$(0, 2)$
Solution
(a) Let circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$ …..$(i)$
Circle $(i)$ cuts orthogonally each of the given three circles.
Then according to the condition
$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}$,
$2g + 17f = c + 4$…..$(ii)$
$7g + 6f = c + 11$…..$(iii)$
$ – g + 22f = c + 3$…..$(iv)$
From $(ii), (iii)$ and $(iv),$ $g = – 3,\,$$f = – 2$
Therefore, the centre of the circle $( – g,\, – f) = \,(3,\,2)$.
Standard 11
Mathematics
