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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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Math Expert
Joined: 02 Sep 2009
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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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05 May 2020, 03:14
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55% (hard)

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62% (02:19) correct 38% (03:37) wrong based on 26 sessions

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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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Joined: 19 Oct 2018
Posts: 1985
Location: India
Re: What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and  [#permalink]

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05 May 2020, 16:28
1
$$x^2 + y^2 + 6x + 8y ≤ 0$$

$$x^2+6x+9-9 +y^2+8y+16-16≤ 0$$

$$(x+3)^2 + (y+4)^2 ≤ 25$$

$$(x+3)^2 + (y+4)^2 ≤ 5^2$$

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 [ 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. $$35π$$

Are You Up For the Challenge: 700 Level Questions
Intern
Joined: 19 Jan 2020
Posts: 44
Re: What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and  [#permalink]

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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?
DS Forum Moderator
Joined: 19 Oct 2018
Posts: 1985
Location: India
Re: What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and  [#permalink]

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28 May 2020, 09:04
1
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.
Intern
Joined: 19 Jan 2020
Posts: 44
Re: What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and  [#permalink]

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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?
DS Forum Moderator
Joined: 19 Oct 2018
Posts: 1985
Location: India
What is the area of the region enclosed by x^2 + y^2 + 6x + 8y ≤ 0 and  [#permalink]

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28 May 2020, 10:03
3
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 [ 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 [ 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 [ 5.85 KiB | Viewed 275 times ]

It's half of the area of the circle.
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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