Question

For the circle x² + y² = r², 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 is maximum –

• A. 4
• B. 5
• C. 6
• D. None of these

Answer 1

Answer: (B)

5

Answer 2

Here, x² + y² = r² and tangents from P(6, 8) are shown as ;

From above figure, in DOMQ, we have

cos q = $\frac{MQ}{OQ}$ and sin q = $\frac{OM}{OQ}$

\ MQ = r cos q and OM = r sin q

\ QR = 2r cos q

PM = OP – OM = 10 – r sin q

\ Area of D PQR = $\frac{1}{2}$ (2r cos q) (10 – r sin q)

\ ƒ(q) = r cos q (10 – r sin q),

{using $\frac{OQ}{OP}$ = sin q ̃ r = 10 sin q}

̃ ƒ(q) = 100 sin q cos q (1 – sin² q) …(i)

ƒ(q) = 100 sin q cos³ q

\ ƒ¢(q) = 100 (cos⁴ q – 3 cos² q sin² q)

ƒ¢¢(q) = 100 (–10 cos³ q sin q + 6 sin³ q cos q)

Put Ģ(q) = 0

̃ tan² q = 1/3 or q = p/6

\ ƒ¢¢(p/6) = 100 $\left( \frac{- 15\sqrt{3}}{8} + \frac{3\sqrt{3}}{8} \right)$ < 0

\ Area is maximum when q = $\frac{\pi}{6}$ and

Hence, r = 10 sin $\frac{\pi}{6}$ = 5.