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If each of the lines 5x+8y=13   and 4x-y=3  contains a diameter of the circle x^{2}+y^{2}-2(a^{2}-7a+11)x-2(a^{2}-6a+6)y+b^{3}+1=0  , then :

Option: 1

a=5\: \: and\: b\not{\varepsilon }(-1,1)


Option: 2

a=-1\: \: and \: \: b\varepsilon (-1 ,1)


Option: 3

a=2\: \: and \: \: b\varepsilon (-\infty ,1)


Option: 4

a=5\: \: and \: \: b\varepsilon (-\infty ,1)


Answers (1)

Given equation of the circle is 

x^{2}+y^{2}-2(a^{2}-7a+11)x-2(a^{2}-6a+6)y+b^{3}+1=0

center of the circle is \left ((a^{2}-7a+11), (a^{2}-6a+6) \right )

The point of intersection of two given lines is (1,1) . Since each of the two given lines contains a diameter of the given circle, therefore the point of intersection of the two given lines is the center of the given circle.

\\a^{2}-7a+11=1\Rightarrow a=2,5\\ a^{2}-6a+6=1\Rightarrow a=1,5\\\text{from above, }a=5

The equation of a given circle becomes

\\x^{2}+y^{2}-2x-2y+b^{3}+1=0\\(x-1)^2+(y-1)^2+b^3-1=0\\b^3=1-(x-1)^2+(y-1)^2\\\therefore b\in(-\infty,1)

Posted by

Sumit Saini

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