Find the area of the region that comprises all points that satisfy the two conditions x2 y2 6x 8y le......

Find the area of the region that comprises all points that satisfy the two conditions x² + y² + 6x + 8y ≤ 0 and 4x ≥ 3y?

x² + y² + 6x + 8y < 0
x² + 6x + 9 – 9 + y² + 8y + 16 – 16 < 0
(x + 3)² + (y + 4)² < 25
This represents a circular region with centre (–3, –4) and radius 5 units.
Substituting x = y = 0, we also see that the inequation is satisfied.
This means that the circle also passes through the origin.
To find out the intercepts that the circle cuts off with the axes, substitute x = 0 to find out the y–intercept and y = 0 to find out x–intercept.
Thus x–intercept = –6 and y–intercept = –8.

Now, the line 4x = 3y passes through the point (–3, –4). Or this line is the diameter of the circle.
The area we are looking for is the area of a semicircle.
Required area = 25 π/2

By: Amit Kumar