send mail to email@example.com mentioning your email id and mobileno registered with us! if details not recieved
Resend Opt after 60 Sec.
Please verify your mobile number
Please update your name
Subscribe to Notifications
Stay updated with the latest Current affairs and other important updates regarding video Lectures, Test Schedules, live sessions etc..
Refer & Earn
My Abhipedia Earning
Kindly Login to view your earning
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
Access to prime resources