Here, x 2 + y 2 = r 2 and tangents from P(6, 8) are shown as ;

From above figure, in Δ OMQ, we have

cos θ = and sin θ =

∴ MQ = r cos θ and OM = r sin θ

∴ QR = 2r cos θ

PM = OP – OM = 10 – r sin θ

∴ Area of Δ PQR = (2r cos θ ) (10 – r sin θ )

∴ ƒ( θ ) = r cos θ (10 – r sin θ ),

{using = sin θ ⇒ r = 10 sin θ }

⇒ ƒ( θ ) = 100 sin θ cos θ (1 – sin 2 θ ) …(i)

ƒ( θ ) = 100 sin θ cos 3 θ

∴ ƒ ′ ( θ ) = 100 (cos 4 θ – 3 cos 2 θ sin 2 θ )

ƒ ′′ ( θ ) = 100 (–10 cos 3 θ sin θ + 6 sin 3 θ cos θ )

Put ƒ ′ ( θ ) = 0

⇒ tan 2 θ = 1/3 or θ = π /6

∴ ƒ ′′ ( π /6) = 100 < 0

∴ Area is maximum when θ = and

Hence, r = 10 sin = 5.