For the circle `x^(2)+y^(2)=r^(2)`, find the value of r for which the area enclosed by the tangents drawn from the point P(6,8) to the circle and the chord of contact and the chord of contact is maximum.

Explanation

The given circle is `x^(2)+y^(2)=r^(2)`. From point (6,8), tangents are drawn to this circle.

Then, the length of tangent is
`PL=sqrt(6^(2)+8^(2)-r^(2))=sqrt(100-r^(2))`
Also, the equation of chord of contact LM is `6x+8y-r^(2)=0`.
`PN=` Length of `_|_` from P to LM
`=(36+64-r^(2))/(sqrt(36+64))=(100-r^(2))/(10)`
Now, in right - angles `Delta PLN`
`LN^(2)=PL^(2)-PN^(2)`
`=(100-r^(2))-((100-r^(2))^(2))/(100)=((100-r^(2))r^(2))/(100)`
or `LN =(r sqrt(100-r^(2)))/(10)`
`:. LM =(rsqrt(100-r^(2)))/(5)` `( :' LM =2LN)`
`:. ` Area of `Delta PLM =Delta =(1)/(2)xxLM xxPN`
`=(1)/(2) xx (r sqrt(100-r^(2)))/(5)xx(100-r^(2))/(10)`
`=(1)/(100)[r(100-r^(2))^(3//2)]`
For the maximum value of `Delta`, we should have
`(dDelta)/(dr)=0`
or `(1)/(100)[(100-r^(2))^(3//2)+r(3)/(2)(100-r^(2))^(1//2)(-2r)]=0`
or `(100-r^(2))^(1//2)[100-r^(2)-3r^(2)]=0`
i.e., `r=10` or `r=5`
But `r=10` gives the length of tangent PL. Therefore, `r cancel (=)10`.
Hence, `r=5`.