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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 is
1. x2 + y2 + 4x - 6y + 4 = 0
2. x2 + y2 + 4x - 6y – 9 = 0
3. x2 + y2 + 4x - 6y - 4 = 0
4. 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.For futhur queries, mail me on navneet25799@gmail. com
Navneet ChandanIIT 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
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