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What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and

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What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and  [#permalink]

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New post 05 May 2020, 03:14
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Re: What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and  [#permalink]

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New post 05 May 2020, 16:28
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\(x^2 + y^2 + 6x + 8y ≤ 0\)

\(x^2+6x+9-9 +y^2+8y+16-16≤ 0\)

\((x+3)^2 + (y+4)^2 ≤ 25\)

\((x+3)^2 + (y+4)^2 ≤ 5^2\)

This represents the region inside the circle whose coordinates of the center are (-3,-4) and whose radius is 5. (Red one)

\(4x ≥ 3y\) represents the region below the line 4x=3y(Blue one). Also, the line 4x=3y passes through the center of the circle, (-3,-4), so it cuts the circle in two equal parts.
Attachment:
Capture.PNG
Capture.PNG [ 40.04 KiB | Viewed 498 times ]




Area of the circle= π*5^2 = 25π

The area of the region enclosed by \(x^2 + y^2 + 6x + 8y ≤ 0\) and \(4x ≥ 3y\) =\( \frac{25π}{2}\)





Bunuel wrote:
What is the area of the region enclosed by \(x^2 + y^2 + 6x + 8y ≤ 0\) and \(4x ≥ 3y\)?


A. \(\frac{25π}{4}\)

B. \(\frac{25π}{2}\)

C. \(25π\)

D. \(30π\)

E. \(35π\)


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Re: What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and  [#permalink]

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New post 28 May 2020, 03:06
nick1816 wrote:
\(x^2 + y^2 + 6x + 8y ≤ 0\)

\(x^2+6x+9-9 +y^2+8y+16-16≤ 0\)

\((x+3)^2 + (y+4)^2 ≤ 25\)

\((x+3)^2 + (y+4)^2 ≤ 5^2\)

This represents the region inside the circle whose coordinates of the center are (-3,-4) and whose radius is 5. (Red one)

\(4x ≥ 3y\) represents the region below the line 4x=3y(Blue one). Also, the line 4x=3y passes through the center of the circle, (-3,-4), so it cuts the circle in two equal parts.
Attachment:
Capture.PNG




Area of the circle= π*5^2 = 25π

The area of the region enclosed by \(x^2 + y^2 + 6x + 8y ≤ 0\) and \(4x ≥ 3y\) =\( \frac{25π}{2}\)





Bunuel wrote:
What is the area of the region enclosed by \(x^2 + y^2 + 6x + 8y ≤ 0\) and \(4x ≥ 3y\)?


A. \(\frac{25π}{4}\)

B. \(\frac{25π}{2}\)

C. \(25π\)

D. \(30π\)

E. \(35π\)


Are You Up For the Challenge: 700 Level Questions



How do we know that is cutting the circle into two equal parts from the equation 4x>=3y.
When we put x=0 then y=0 and the line passes from origin. But how does it divide the circle?
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Re: What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and  [#permalink]

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New post 28 May 2020, 09:04
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nikitamaheshwari

4*(-3) = 3*(-4)

so, line 4x=3y passes through (-3,-4) point, that is the center of the circle. So basically this line is one of the diameters of the circle; Hence, it'll cut the circle in 2 equal parts.
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Re: What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and  [#permalink]

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New post 28 May 2020, 09:10
nick1816 wrote:
nikitamaheshwari

4*(-3) = 3*(-4)

so, line 4x=3y passes through (-3,-4) point, that is the center of the circle. So basically this line is one of the diameters of the circle; Hence, it'll cut the circle in 2 equal parts.



Thanks. But this is true when we consider 4x=3y. if we take any value like 4(5)> 3(3). then we dont get to know whether it will lie on the circle. right?
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What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and  [#permalink]

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New post 28 May 2020, 10:03
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nikitamaheshwari
Now i can see where you're getting confused

y≤ mx represents the region below the line y=mx (including points on the line).

y ≥ mx represents the region below the line y=mx (including points on the line).

So \(y≤ \frac{4x}{3}\) represents the region(in grey) below the line \(y=\frac{4x}{3}\) (in red).
Attachment:
Untitled.png
Untitled.png [ 5.19 KiB | Viewed 277 times ]


Also, \(x^2+y^2≤ r^2\) represents the region inside the circle (including points on the circle).

\(x^2+y^2 ≥ r^2\) represents the region outside the circle (including points on the circle).

So, \((x+3)^2+(y+4)^2 ≤ 5^2\) represents the region (in grey) inside the circle \((x+3)^2+(y+4)^2 = 5^2\) (in red)
Attachment:
Untitled 1.png
Untitled 1.png [ 5.95 KiB | Viewed 276 times ]


Now we have to find the common region (in grey), which is below the line y= 4x/3 and which is inside the circle.
Attachment:
Untitled.png
Untitled.png [ 5.85 KiB | Viewed 275 times ]


It's half of the area of the circle.
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What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and   [#permalink] 28 May 2020, 10:03

What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and

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