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The equation of the tangent to the circle r = 2a cosθ at any point (r1, θ1) is given by-
- a)r cos (θ-θ1)=2a cos2θ1
- b)r cos (θ-2θ1)=a cos2θ1
- c)r cos (θ-θ1)= cos2θ1
- d)r cos (θ-2θ1)=2a cos2θ1
Correct answer is option 'D'. Can you explain this answer?
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The equation of the tangent to the circle r =2a cosθ at any poin...
To find the equation of the tangent to the circle with equation r = 2a cosθ, we need to find the slope of the tangent line at a given point on the circle.
The equation of a circle in polar coordinates is r = a cosθ, where r is the radius, a is the radius of the circle, and θ is the angle.
To find the slope of the tangent line, we can take the derivative of r with respect to θ:
dr/dθ = -a sinθ
The slope of the tangent line is given by the derivative dr/dθ.
Let's assume we want to find the equation of the tangent line at a specific point (r, θ) on the circle.
The slope of the tangent line at that point is given by dr/dθ evaluated at that point:
slope = dr/dθ = -a sinθ
Now, to find the equation of the tangent line, we use the point-slope form:
y - y1 = m(x - x1)
where (x1, y1) is the point on the circle and m is the slope of the tangent line.
In polar coordinates, the coordinates (x, y) are related to (r, θ) by:
x = r cosθ
y = r sinθ
Substituting these values into the point-slope form, we get:
y - (r sinθ) = (-a sinθ)(x - r cosθ)
Simplifying, we have:
y - r sinθ = -a sinθ(x - r cosθ)
This is the equation of the tangent line to the circle r = 2a cosθ at the point (r, θ).
The equation of a circle in polar coordinates is r = a cosθ, where r is the radius, a is the radius of the circle, and θ is the angle.
To find the slope of the tangent line, we can take the derivative of r with respect to θ:
dr/dθ = -a sinθ
The slope of the tangent line is given by the derivative dr/dθ.
Let's assume we want to find the equation of the tangent line at a specific point (r, θ) on the circle.
The slope of the tangent line at that point is given by dr/dθ evaluated at that point:
slope = dr/dθ = -a sinθ
Now, to find the equation of the tangent line, we use the point-slope form:
y - y1 = m(x - x1)
where (x1, y1) is the point on the circle and m is the slope of the tangent line.
In polar coordinates, the coordinates (x, y) are related to (r, θ) by:
x = r cosθ
y = r sinθ
Substituting these values into the point-slope form, we get:
y - (r sinθ) = (-a sinθ)(x - r cosθ)
Simplifying, we have:
y - r sinθ = -a sinθ(x - r cosθ)
This is the equation of the tangent line to the circle r = 2a cosθ at the point (r, θ).
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