A circle of radius 1 unit touches the positive x-axis and the positive y-axis at `Aa n dB` , respectively. A variable line passing through the origin intersects the circle at two points `Da n dE` . If the area of triangle `D E B` is maximum when the slope of the line is `m ,` then find the value of `m^(-2)`

Explanation

The equation of the circle is
`(x-1)^(2)+(y-1)^(2)=1`
or `x^(2)+y^(2)-2x-2y+1=0` (1)
Let the equation of the variable straight line through origin be
`y=mx` (2)
This line intersects the cricle at two points D and E.
Distance of line from point B(0,1) is
`BM=(|m(0)-1|)/(sqrt(1+m^(2)))=(1)/(sqrt(1+m^(2)))`

Also, `CP=(|m(1)-1|)/(sqrt(1+m^(2)))=(|m-1|)/(sqrt(1+m^(2)))`
`:. DE=2PE =2sqrt(CE^(2)-CP^(2))=2sqrt(1-((m-1)^(2))/(1+m^(2)))=2sqrt((2m)/(1+m^(2)))`
`:. `Area of triangle DEB.
`Delta =(1)/(2) BM xx DE`
`=(1)/(2) (1)/(sqrt(1+m^(2)))xx2sqrt((2m)/(1+m^(2)))=(sqrt(2m))/(1+m^(2))`
Differentiating `Delta` w.r.t. m, we get
`(d Delta)/(dm)=sqrt(2)[((1)/(2sqrt(m))(1+m^(2))-2msqrt(m))/((1+m^(2))^(2))]=(1-3m^(2))/(sqrt(2)sqrt(m)(1+m^(2))^(2))`
If `(d Delta)/(dm)=0` , then `m=(1)/(sqrt(m))`, whichi is point of maxima.
Therefore, area is maximum if `m=(1)/(sqrt(3))`.