The line `3x+6y=k` intersects the curve `2x^2+3y^2=1` at points `A and B` . The circle on `A B` as diameter passes through the origin. Then the value of `k^2` is__________

Answer

9

Explanation

`3x+6y=k`
or `(3x+6y)/(k)=1 ` (1)
Also, `2x^(2)+2xy+3y^(2)-1=0` (2)
Now, homogenizing (2) with the helop of (1), we get
`2x^(2)+2xy+3y^(2)-((3x+6y)/(k))^(2)=0`

or `k^(2)(2x^(2)+2xy+3y^(2))-(3x+6y)^(2)=0` (3)
This is the equation of the pair of straight lines OA and OB. ltbr. Now, AB is diameter. Then OA and OB are perpendicular . So, Coefficient of `x^(2)+` Coefficient of `y^(2)=0`
or `(2k^(2) -9)+(3k^(2)-36) =0`
or `5k^(2)=45`
or `k^(2) =9`