The circle is symmetric with respect to the x and y axes, hence we can find the area of one quarter of a circle and multiply by 4 in order to obtain the total area of the circle.
Solve the above equation for \( y \)
\( y = \pm \sqrt{ a^2 - x^2 } \)
The equation of the upper semi circle (y positive) is given by
\( y = \sqrt { a^2 - x^2 } \)
Factor out \( a^2 \) inside the radicand
\( y = \sqrt { a^2(1 - x^2/a^2) } \)
Take \( a^2 \) from under the radicand and rewrite \( y \) as follows
\( y = a \sqrt { 1 - x^2 / a^2 } \)
We use integrals to find the area of the upper right quarter of the circle as follows
(1 / 4) Area of circle = \( \displaystyle \int_0^a a \sqrt{1-x^2/a^2} dx \)
Let us substitute \( \; x / a \) by \( \; \sin t \) so that \( \sin t = x / a \) and \( dx = a \cos t \; dt \; \) and the area is given by
(1 / 4) Area of circle = \( \displaystyle \int_0^{\pi/2} a^2 \sqrt{1-\sin^2t} \cos t \; dt\)
We now use the trigonometric identity
\( \sin^2 t + \cos^2 t = 1 \)
which gives
\( \sqrt{1-\sin^2t} = \cos t \quad \) since t varies from 0 to \( \pi/2 \) hence
(1 / 4) Area of circle = \( \displaystyle \int_0^{\pi/2} a^2 \cos^2t \; dt\)
Use the trigonometric identity \( \; \cos^2 t = ( \cos 2t + 1 ) / 2 \;\) to linearize the integrand;
(1 / 4) Area of circle = \( \displaystyle \int_0^{\pi/2} a^2 ( \cos 2t + 1 ) / 2 \; dt\)
Evaluate the integral
(1 / 4) Area of circle = \( \displaystyle (1/2)a^2 \left[(1/2) \sin 2t + t\right]_0^{\pi/2} \)
Simplify
(1 / 4) Area of circle = \( (1/4) \pi a^2 \)
The total area of the circle is obtained by a multiplication by 4
Area of circle = \( 4 \times (1/4) \pi a^2 = \pi a^2 \)