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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. Then the equation of the circle with AB as its diameter is

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
A point on the hyperbola is `(3 sec theta, 2 tan theta)`.
It lies on the circle. S, `9 sec^(2) theta+4 tan^(2) theta-24 sec theta=0`.
`"i.e., "13 sec^(2) theta-24 sec theta-4=0`
`"or "sec theta =2,-(2)/(13)`
Clearly circle cuts the hyperbola in first and fourth quadrants.
`therefore" "sec theta=2, tan theta=sqrt3`
The points of intersection are `A(6,2sqrt3) and B(6,-2sqrt3)`, ltBrgt Therefore, the circle with AB as diameter is
`(x-6)^(2)+y^(2)=(2sqrt3)^(2)`
`"or "x^(2)+y^(2)-12x+24=0`
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