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Find the area of the region that comprises all points that satisfy the two conditions x2 + y2 + 6x + 8y ≤ 0 and 4x ≥ 3y?
None of these
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 ProfileResourcesReport error
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