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

Answer 1

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

1. xcosθ + ysinθ = 5
2. −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 + sin⁡2θ, which varies from 0 to 2, we get:

x² + y² = 50

which represents a circle of radius `sqrt50 = 5sqrt5`​ centered at the origin.