Tangents are drawn to the circle x2+y2=50 from a point 'P' lying on the x-axis. These tangents meet the y-axis at points 'P1' and 'P2' . Possible coordinates of 'P' so that area of triangle PP1P2 is minimum, are

Question

Tangents are drawn to the circle x2+y2=50 from a point 'P' lying on the x-axis. These tangents meet the y-axis at points 'P1' and 'P2' . Possible coordinates of 'P' so that area of triangle PP1P2 is minimum, are (A) (10 ,0) (B) (5 √2 , 0) (C) (5 ,0) (D) ((5/√2) , 0)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-kxahi3xubqqplhy6.jpg" /> OP=5√2sec θ OP1=5√2cosec θ area (Δ P P1 P2)=(100/s i n 2 θ ) area (Δ P P1 P2)m i n=100⇒ θ =π /4⇒ OP=10 P(10 , 0)⇒ or P(- 10 ,0) Hence (A) is correct

Q. Tangents are drawn to the circle $x^{2} + y^{2} = 50$ from a point $^{'} P^{'}$ lying on the x-axis. These tangents meet the y-axis at points $^{'} P_{1}^{'}$ and $^{'} P_{2}^{'}$ . Possible coordinates of $^{'} P^{'}$ so that area of triangle $P P_{1} P_{2}$ is minimum, are

$(10, 0)$

$(52, 0)$

$(5, 0)$

$(\frac{5}{2}, 0)$

$OP = 52 sec θ$
$O P_{1} = 52 cosec θ$
area $(Δ P P_{1} P_{2}) = \frac{100}{s in 2 θ}$
area $(Δ P P_{1} P_{2})_{min} = 100 \Rightarrow θ = π /4 \Rightarrow OP = 10$
$P (10, 0) \Rightarrow$ or $P (- 10, 0)$
Hence $(A)$ is correct

Q. Tangents are drawn to the circle x2+y2=50 from a point ′P′ lying on the x-axis. These tangents meet the y-axis at points ′P1′​ and ′P2′​ . Possible coordinates of ′P′ so that area of triangle PP1​P2​ is minimum, are

(10,0)

(52​,0)

(5,0)

(2​5​,0)