The ellipse `4x^2+9y^2=36` and the hyperbola `a^2x^2-y^2=4` intersect at right angles. Then the equation of the circle through the points of intersection of two conics is (a) `x^2+y^2=5` (b) `sqrt(5)(x^2+y^2)-3x-4y=0` (c) `sqrt(5)(x^2+y^2)+3x+4y=0` (d) `x^2+y^2=25`

Options

A. `x^(2)+y^(2)=5`
B. `sqrt5(x^(2)+y^(2))-3x-4y=0`
C. `sqrt5(x^(2)+y^(2))+3x+4y=0`
D. `x^(2)+y^(2)=25`

Answer

A

Explanation

Since ellipse and hyperbola intersect orthohonally, they are confocal.
Hence, a= 2 (equating foci).
Let the point of intersection in the first quadrant be `P(x_(1),y_(1))`.
P lies on both the curves. Therefore,
`4x_(1)^(2)+9y_(1)^(2)=36 and 4x_(1)^(2)-y_(1)^(2)=4`
Adding these two results, we get
`8(x_(1)^(2)+y_(1)^(2))=40`
`"or "x_(1)^(2)+y_(1)^(2)=5`
`"or "r=sqrt5`
Hence, the equation of the circle is
`x^(2)+y^(2)=5`