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10-1.Circle and System of Circles
hard
The equation of the circle which intersects circles ${x^2} + {y^2} + x + 2y + 3 = 0$, ${x^2} + {y^2} + 2x + 4y + 5 = 0$and ${x^2} + {y^2} - 7x - 8y - 9 = 0$ at right angle, will be
A
${x^2} + {y^2} - 4x - 4y - 3 = 0$
B
$3({x^2} + {y^2}) + 4x - 4y - 3 = 0$
C
${x^2} + {y^2} + 4x + 4y - 3 = 0$
D
$3({x^2} + {y^2}) + 4(x + y) - 3 = 0$
Solution
(d) Let circle be ${x^2} + {y^2} + 2gx + 2fy + c = 0$.
Then according to the conditions given,
$g + 2f = c + 3$….$(i)$
$2g + 4f = c + 5$….$(ii)$
$ – 7g – 8f = c – 9$….$(iii)$
$ \Rightarrow g = \frac{2}{3},\;f = \frac{2}{3},\;c = – 1$
Therefore, the required equation is
$3({x^2} + {y^2}) + 4(x + y) – 3 = 0$.
Standard 11
Mathematics
