Let `2x^2 +y^2-3xy = 0` be the equation of a pair of tangents drawn from the origin `O` to a circle of radius `3` with centre in the first quadrant. If `A` is one of the points of contact, then the length of `OA` is

Explanation

`2x^(2)+y^(2)-3xy=0" "["given"]`
`rArr 2x^(2)-2xy-xy+y^(2)=0`
`rArr2x(x-y)-y(x-y)=0`
`rArr(2x-y)(x-y)=0`
`rArr y=2x,y=x`
are the equations of straight lines passing through origin.
Now, let the angle between the lines be 2 ` theta` and the line y = x makes angle of `45^(@)` with X-axis.
Therefore, `tan(45^(@)+2 theta) = 2 ` (slope of the line y = 2x)

`rArr (tan45^(@)+tan2theta)/(1-tan45^(@)xxtan2theta)=2rArr(1+tan2theta)/(1-tan2theta)=2`
`rArr ((1+tan2theta)-(1-tan2theta))/((1+tan2theta)+(1-tan2theta))=(2-1)/(2+1)=(1)/(3)`
`rArr (2tan2theta)/(2)=(1)/(3)rArrtan2theta=(1)/(3)`
`rArr (2tantheta)/(1-tan^(2)theta)=(1)/(3)`
`rArr (2tantheta)*3=1-tan^(2)theta`
`rArr tan^(2)theta+6tantheta-1=0`
`rArr tantheta=(-6pmsqrt(36+4xx1xx1))/(2)=(-6pmsqrt40)/(2)`
`rArr tantheta=-2pmsqrt10" "[therefore0ltthetalt(pi)/(4)]`
`rArr tantheta=-3+sqrt10`
Again, in `DeltaOCA`
`tantheta = (3)/(OA)`
`therefore OA=(3)/(tantheta)=(3)/(-3+sqrt10)=(3(3+sqrt10))/((-3+sqrt10)(3+sqrt10))`
`=(3(3+sqrt10))/((10-9))=3(3+sqrt10)`