QUESTON 1).The angle between a pair of tangents drawn from a point P to the circle x2 + y2 + 4x - 6y + 9 sin2 + 13cos2 = 0 is 2. The equation of the locus of the point P is1. x2 + y2 + 4x - 6y + 4 = 02. x2 + y2 + 4x - 6y – 9 = 03. x2 + y2 + 4x - 6y - 4 = 04. x2 + y2 + 4x - 6y + 9 = 0

The equation of the circle is x² + y² + 4x - 6y + 9 sin² +13 cos ² = 0

Hence, the center of the circle = (-2, 3)

Radius of the circle is v (-2)² + (3²) – 9 sin² -13cos²

= v13 – 9 sin² -13cos²

= v13sin² – 9 sin²

= v4 sin² = 2 sin

Now, sin = OA/OP = 2 sin / v(h+2)² + (k-3)²

Hence, (h+2)² + (k-3)² = 4

h² + k² + 4h -6k +9 = 0.

Therefore, the locus of P is x² +y² + 4x - 6y +9 = 0.

image related to the above solution.
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Navneet Chandan
IIT DELHI

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