From the point p (-1,-2) pq and pr are the tangent through a circle
Answers
Given : From Point P ( - 1 , - 2) , PQ & PR are tangent drawn to Circle x² + y² - 6x - 8y = 0 .
To Find : The Angle subtended by QR at the center of the circle
Solution:
P ( - 1 , -2)
Circle x² + y² - 6x - 8y = 0
=> (x - 3)² - 9 + (y - 4)² - 16 = 0
=> (x - 3)² + (y - 4)² = 5²
Center = ( 3 , 4) radius = 5
O = center of circle
PO = √(3 - (-1))² +(4 -(-2))² = √4² + 6² = √52 = 2√13
Cos ∠POQ = OQ/OP
=> Cos ∠POQ = 5/2√13
∠POQ = Cos⁻¹ (5/2√13)
∠QOR = 2 ∠POQ = 2 Cos⁻¹ (5/2√13)
Using Cos⁻¹x + Sin⁻¹x = π/2
= 2 (π/2 - Sin⁻¹ (5/2√13))
= π - 2 Sin⁻¹ (5/2√13)
Angle Subtended by QR at the center of the Circle = π - 2 Sin⁻¹ (5/2√13)
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