Question

Find the perimeter of the circle x^2 +y^2=a^2

Answer 1

x^2 + y^2 = a^2 is a circle with centre (0,0) and radius a units.

To find the perimeter of this circle let us find the arc length of the quarter circle and then multiply answer with 4.

x^2 + y^2 = a^2. Differentiate w.r.t.x

2x + 2y dy/dx =0

dy/dx = -2x/2y = -x/y.

√{1+(dy/dx)^2} = √{1+(-x/y)^2} =√{(y^2+x^2)/y^2}

=√(a^2/y^2) since x^2 +y^2 = a^2.

=a/y = a/√(a^2 - x^2)

So length of the arc of the quarter circle

= integ.0 to a √{1+(dy/dx)^2} dx

=integ. 0 to a a/√(a^2 - x^2) dx

=limit 0 to a a.sin^(-1) x/a

= a.{ sin^(-1) 1 - sin^(-1) 0}

=a× π/2.

So perimeter of the circle = 4× πa/2

=2πa units.

Answer 2

Answer: Circumference or perimeter of circle is $2\pi a$  units .

Step-by-step explanation:

Concept : Equation of Circles

Given : $x^{2} + y^{2} = a^{2}$

To Find : Perimeter of the given circle

Solution :

The given equation of circle is $x^{2} + y^{2} = a^{2}$

where the co - ordinates are (0, 0) and radius = a units

∴ Perimeter of the circle = $2\pi (radius)$

                                         = $2 \pi a$ units.

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