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Question
Answer the following:
Find the equation of locus of the point of intersection of perpendicular tangents drawn to the circle x = 5 cos θ and y = 5 sin θ
Solution
x = 5cosθ, y = 5sinθ
Step 1: General Equation of a Tangent to a Circle
x = Rcosθ, y = Rsinθ
xcosθ + ysinθ = R
For our given circle of radius R = 5, the equation of a tangent is:
xcosθ + ysinθ = 5
Step 2: Equation of Perpendicular Tangents
Using trigonometric identities:
cos(θ + 90∘) = −sinθ, sin(θ + 90∘) = cosθ
x(−sinθ) + y(cosθ) = 5
−xsinθ + ycosθ = 5
Step 3: Solving for the Intersection of the Tangents
- xcosθ + ysinθ = 5
- −xsinθ + ycosθ = 5
Multiply the first equation by cos θ and the second equation by sin θ:
xcos2θ + ysinθcosθ = 5cosθ
−xsinθcosθ + ycos2θ = 5sinθ
x(cos2θ − sin2θ) + y(2sinθcosθ) = 5cosθ + 5sinθ
cos2θ − sin2θ = cos2θ, 2sinθcosθ = sin2θ
xcos2θ + ysin2θ = 5(cosθ + sinθ)
x2 + y2 = 52(1 + 2cosθsinθ)
x2 + y2 = 25(1 + sin2θ)
Taking the maximum value of 1 + sin2θ, which varies from 0 to 2, we get:
x2 + y2 = 50
which represents a circle of radius `sqrt50 = 5sqrt5` centered at the origin.
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