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The given circle is `x^(2)+y^(2)=r^(2)`. From point (6,8), tangents are drawn to this circle. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CEN_CG_C04_S01_113_S01.png" width="80%"> <br> Then, the length of tangent is <br> `PL=sqrt(6^(2)+8^(2)-r^(2))=sqrt(100-r^(2))` <br> Also, the equation of chord of contact LM is `6x+8y-r^(2)=0`. <br> `PN=` Length of `_|_` from P to LM <br> `=(36+64-r^(2))/(sqrt(36+64))=(100-r^(2))/(10)` <br> Now, in right - angles `Delta PLN` <br> `LN^(2)=PL^(2)-PN^(2)` <br> `=(100-r^(2))-((100-r^(2))^(2))/(100)=((100-r^(2))r^(2))/(100)` <br> or `LN =(r sqrt(100-r^(2)))/(10)` <br> `:. LM =(rsqrt(100-r^(2)))/(5)` `( :' LM =2LN)` <br> `:. ` Area of `Delta PLM =Delta =(1)/(2)xxLM xxPN` <br> `=(1)/(2) xx (r sqrt(100-r^(2)))/(5)xx(100-r^(2))/(10)` <br> `=(1)/(100)[r(100-r^(2))^(3//2)]` <br> For the maximum value of `Delta`, we should have <br> `(dDelta)/(dr)=0` <br> or `(1)/(100)[(100-r^(2))^(3//2)+r(3)/(2)(100-r^(2))^(1//2)(-2r)]=0` <br> or `(100-r^(2))^(1//2)[100-r^(2)-3r^(2)]=0` <br> i.e., `r=10` or `r=5` <br> But `r=10` gives the length of tangent PL. Therefore, `r cancel (=)10`. <br> Hence, `r=5`.