How to solve this optimization type question?

You need a line that is tangent to the circle. The point of tangency is in the first quadrant, so x > 0 and y > 0 (well, greater than or equal to...). Using the x-axis, the y-axis and the tangent line, you'll have a right triangle. Find the area of the triangle

x^2 + y^2 = 1

Tangent line? Find the derivative

2x * dx + 2y * dy = 0

x * dx + y * dy = 0

y * dy = -x * dx

dy/dx = -x/y

dy/dx = -x / sqrt(1 - x^2)

Let x = a and y will equal sqrt(1 - a^2)

We need a line with a slope of -a / sqrt(1 - a^2) that passes through (a , sqrt(1 - a^2)). The reason we're using a different variable is just because using x and y for different things will get confusing.

y - yo = m * (x - xo)

y - sqrt(1 - a^2) = (-a/sqrt(1 - a^2)) * (x - a)

y - sqrt(1 - a^2) = (a/sqrt(1 - a^2)) * (a - x)

y = sqrt(1 - a^2) + (a / sqrt(1 - a^2)) * (a - x)

Now, find the intercepts.

x-intercept happens when y = 0

0 = sqrt(1 - a^2),+ (a / sqrt(1 - a^2)) * (a - x)

(a / sqrt(1 - a^2)) * (x - a) = sqrt(1 - a^2)

a * (x - a) = (1 - a^2)

ax - a^2 = 1 - a^2

ax = 1

x = 1/a

(1/a , 0) is the x-intercept

y-intercept happens when x = 0

y = sqrt(1 - a^2) + (a / sqrt(1 - a^2)) * (a - x)

y = sqrt(1 - a^2) + (a / sqrt(1 - a^2)) * (a - 0)

y = sqrt(1 - a^2) + a^2 / sqrt(1 - a^2)

y = (1 - a^2 + a^2) / sqrt(1 - a^2)

y = 1 / sqrt(1 - a^2)

(0 , 1 / sqrt(1 - a^2)) is the y-intercept

Now your right triangle has 2 side lengths of 1/a and 1/sqrt(1 - a^2)

A = (1/2) * (1/a) * (1/sqrt(1 - a^2))

Find dA/da and set it to 0. This is when your change in area minimizes or maximizes

A = 1 / (2a * sqrt(1 - a^2))

A = 1/(2 * sqrt(a^2 - a^4))

A^2 = 1 / (4 * (a^2 - a^4))

A^(-2) = 4 * (a^2 - a^4)

-2 * A^(-3) * dA/da = 4 * (2a - 4a^3)

dA/da = 0

-2 * A^(-3) * 0 = 4 * (2a - 4a^3)

0 = 4 * (2a - 4a^3)

0 = 2a - 4a^3

0 = a - 2a^3

0 = a * (1 - 2a^2)

a = 0

1 - 2a^2 = 0

2a^2 = 1

a^2 = 1/2

a = +/- sqrt(2)/2

Since a > 0, then a = sqrt(2)/2

(sqrt(2)/2)^2 + y^2 = 1

1/2 + y^2 = 1

y^2 = 1/2

y = +/- sqrt(2)/2

y > 0 as well

(sqrt(2)/2 , sqrt(2)/2) is the point