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The circle x^2+y^2-8x=0 and hyperbola ...

The circle `x^2+y^2-8x=0` and hyperbola `x^2/9-y^2/4=1` intersect at the points A and B Equation of the circle with AB as its diameter is

A

`x^(2) +y^(2)-12x+24=0`

B

`x^(2) +y^(2)+12x+24=0`

C

`x^(2) +y^(2)+24x-12=0`

D

`x^(2) +y^(2)-24x-12=0`

Text Solution

Verified by Experts

The equation of the hyperbola is`(x^(2))/(9)-(y^(2))/(4)=1 ` and that of circle is `x^(2) +y^(2)-8x=0`
For their points of intersection, `(x^(2))/(9)+(x^(2)-8X)/(4)=1`
` rArr 4x^(2) +9x^(2)-72 x =36`
`rArr 13x^(2)-72 x -36=0`
` rArr 13x^(2)-78x+6x-36=0`
`rArr 13x(x-6)+6(x-6)=0`
`rArr x=6, x=-(13)/(6)`
`x= -(13)/(6)` not acceptable.
Now, for `x=6, y=pm 2sqrt(3)`
Required equation is, ` (x-6)^(2)+(y+2sqrt(3))(y-2sqrt(3))=0`
`rArr x^(2)-12x+y^(2)+24=0`
`x^(2)+y^(2)-12x+24=0`
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