Math, asked by thangamba22051, 11 months ago

The angle between a pair of tangents drawn from a point P to the circle x² + y² + 4x – 6y + 9 sin²α + 13 cos²α = 0 is 2α. The equation of the locus of the point P is
(a) x² + y² + 4x – 6y + 4 = 0
(b) x² + y² + 4x – 6y – 9 = 0
(c) x² + y² + 4x – 6y – 4 = 0
(d) x² + y² + 4x – 6y + 9 = 0

Answers

Answered by sks2971996
11

Ans is D. I hope this will help you.

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Answered by amitnrw
5

x² + y² + 4x - 6y + 9 = 0 is the Locus of point P if The angle between a pair of tangents drawn from a point P to the circle x² + y² + 4x – 6y + 9 sin²α + 13 cos²α = 0 is 2α.

Step-by-step explanation:

x² + y² + 4x – 6y + 9 sin²α + 13 cos²α = 0

=> x² + 4x + 4 - 4  + y² - 6y  + 9  - 9 + 9 sin²α + 13 cos²α = 0

=> (x +2)² +(y - 3)² - 13 + 9 sin²α + 13 cos²α = 0

=>  (x +2)² +(y - 3)² = 13 - 13 cos²α - 9 sin²α

=> (x +2)² +(y - 3)² =  13Sin²α - 9 sin²α

=>  (x +2)² +(y - 3)² =  4Sin²α

=>   (x +2)² +(y - 3)² =  (2Sinα)²

center = -2 , 3  

radius = 2 Sinα

The angle between a pair of tangents drawn from a point P = 2α

half of 2α = α

=> Sinα = Radius/(distance of P from center of Circle)

=> distance of P from center of Circle = Radius/Sinα

=> distance of P from center of Circle = 2 Sinα/Sinα

=>  distance of P from center of Circle = 2

Distance of P from center

= √(x - (-2))² + (y - 3)²  = 2

Squaring both sides

=> x² + 4x + 4 + y² - 6y + 9 = 4

=> x² + y² + 4x - 6y + 9 = 0

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