SOLUTION: Find the area of the region enclosed by the graph of the equation $x^2-14x+3y+70=15+9y-y^2$ that lies below the line $y=x-3$.

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Find the area of the region enclosed by the graph of the equation $x^2-14x+3y+70=15+9y-y^2$ that lies below the line $y=x-3$.      Log On


   

Question 1203378: Find the area of the region enclosed by the graph of the equation $x^2-14x+3y+70=15+9y-y^2$ that lies below the line $y=x-3$.
Found 3 solutions by Alan3354, Edwin McCravy, math_tutor2020:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the area of the region enclosed by the graph of the equation $x^2-14x+3y+70=15+9y-y^2$ that lies below the line $y=x-3$.
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Find the area of the region enclosed by the graph of the equation x^2-14x+3y+70=15+9y-y^2 that lies below the line y=x-3.
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x^2-14x+3y+70=15+9y-y^2 is a circle. Find the center and the radius:
x^2-14x+49 + y^2-6y+9 = 15 - 70 + 58 = 3
--> (x-7)^2 + (y-3)^2 = 3
Center at (7,3): r = sqrt(3)
-----
Find the distance from the center to the line x-y - 3 = 0
d = |1 - 1 -3|/sqrt(1+1) = 3/sqrt(2) = 3sqrt(2)/2
-----
Find the area of the minor segment, i.e., the area above the given line:
h = distance from the line to the center of the segment
h = 3 - 3sqrt(2)/2
Area = r^2*acos(1-h/r) - (r-h)*sqrt(2rh - h^2) ---- (angle in radians)
-----------------
h = ~ 3 - 2.12132 = 0.87868
Sub and find the area of the segment, then subtract that from the area of the circle.

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
First we get the equation x^2-14x+3y+70=15+9y-y^2 into standard form:

x%5E2-14x%2B3y%2B70=15%2B9y-y%5E2
x%5E2-14x%2By%5E2%2B3y-9y=15-70
x%5E2-14x%2By%5E2-6y=-55
completing the square:
%28x%5E2-14x%2B49%29%2B%28y%5E2-6y%2B9%29=-55%2B49%2B9
%28x-7%29%5E2%2B%28y-3%29%5E2=3
So it's a circle with center (7,3) and radius sqrt%283%29

Now we'll find the points of intersection with the line y = x-3:

x%5E2-14x%2By%5E2-6y=-55
x%5E2-14x%2B%28x-3%29%5E2-6%28x-3%29=-55
x%5E2-14x%2B%28x%5E2-6x%2B9%29-6x%2B18=-55
x%5E2-14x%2Bx%5E2-6x%2B9-6x%2B18=-55
2x%5E2-26x%2B27=-55
2x%5E2-26x%2B82=0
x+=+%28-%28-26%29+%2B-+sqrt%28%28-26%29%5E2-4%2A2%2A82+%29%29%2F%282%2A2%29+
x+=+%2826+%2B-+sqrt%28676-656%29%29%2F%284%29+
x+=+%2826+%2B-+sqrt%2820%29%29%2F%284%29+
x+=+%2826+%2B-+sqrt%284%2A5%29%29%2F%284%29+
x+=+%2826+%2B-+2sqrt%285%29%29%2F%284%29+
x+=+%282%2813+%2B-+sqrt%285%29%29%29%2F%284%29+
x+=+%2813+%2B-+sqrt%285%29%29%2F%282%29+

y=x-3
y=%2813+%2B-+sqrt%285%29%29%2F%282%29-3 
y=%2813+%2B-+sqrt%285%29%29%2F%282%29-6%2F2
y=%287+%2B-+sqrt%285%29%29%2F%282%29
So the points of intersection are 
Let's graph the circle and line:

 


We want to find the major segment of the circle below the line.
We now see that calculus is not the way to go here.  So we must
resort to geometry and analytic geometry.  We draw a radius and
a perpendicular to the chord, which forms a right triangle 

 

We find the length of the green side of the right triangle. One end is
the center of the circle (7,3).  The other end is the midpoint of the
chord.  We find the midpoint of the chord:

expr%281%2F2%29%28x%5B1%5D%2Bx%5B2%5D%29%22%22=%22%22expr%281%2F2%29%28%2813+-+sqrt%285%29%29%2F%282%29%2B%2813+%2B+sqrt%285%29%29%2F%282%29%29%22%22=%22%22expr%281%2F2%29%2826%2F2%29%22%22=%22%2213%2F2

expr%281%2F2%29%28y%5B1%5D%2By%5B2%5D%29%22%22=%22%22expr%281%2F2%29%28%287+-+sqrt%285%29%29%2F%282%29%2B%287+%2B+sqrt%285%29%29%2F%282%29%29%22%22=%22%22expr%281%2F2%29%2814%2F2%29%22%22=%22%227%2F2

So the midpoint of the chord is %28matrix%281%2C3%2C13%2F2%2C%22%2C%22%2C7%2F2%29%29

We use the distance formula to find the length of the green side

sqrt%28%287-13%2F2%29%5E2%2B%283-7%2F2%29%5E2%29%22%22=%22%22sqrt%28%2814%2F2-13%2F2%29%5E2%2B%286%2F2-7%2F2%29%5E2%29%22%22=%22%22sqrt%28%281%2F2%29%5E2%2B%28-1%2F2%29%5E2%29%22%22=%22%22sqrt%281%2F4%2B1%2F4%29%22%22=%22%22sqrt%282%2F4%29%22%22=%22%22sqrt%282%29%2F2  

So the green leg of the right triangle is sqrt%282%29%2F2 

We use the Pythagorean theorem to find the other leg of the right triangle,
The hypotenuse is the radius of the circle sqrt%283%29.

other leg = half of chord = sqrt%28sqrt%283%29%5E2-%28sqrt%282%29%2F2%29%5E2%29%22%22=%22%22sqrt%283-2%2F4%29%22%22=%22%22sqrt%286%2F2-1%2F2%29%22%22=%22%22sqrt%285%2F2%29%22%22=%22%22sqrt%28%285%2A2%29%2F%282%2A2%29%29%22%22=%22%22sqrt%2810%29%2F2

So the area of the right triangle is

expr%281%2F2%29%28base%29%28height%29%22%22=%22%22expr%281%2F2%29%2A%28sqrt%2810%29%2F2%29%2A%28sqrt%282%29%2F2%29sqrt%2820%29%2F8%22%22=%22%22sqrt%284%2A5%29%2F8%22%22=%22%222sqrt%285%29%2F8%22%22=%22%22sqrt%285%29%2F4

We find the angle between the red and green lines from

cos%28angle%29=%28sqrt%282%29%29%2F2%29%2F%28sqrt%283%29%29=0.4082482905 
angle=1.150261991radians


We now draw in the radius to the other point of intersection of the circle and the line:

 

This gives us another right triangle congruent to the other one.
Now the angle on the minor segment is twice the angle above or

2%281.150261991%29radians or 2.300523982radians

To find the angle of the major segment we subtract from 2pi

2pi-2.300523982radians%22%22=%22%223.982661325radians

The area of the sector = expr%281%2F2%29%28radius%29%28matrix%281%2C3%2Cangle%2Cin%2Cradians%29%29%22%22=%22%22

expr%281%2F2%29%28sqrt%283%29%5E%22%22%29%283.982661325%29%22%22=%22%223.449085882

Finally we add the area of the two right triangles:

3.449085882%2B2%2A%28sqrt%285%29%2F4%29%22%22=%22%224.567119871 square units

Edwin

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

I'll give the overall outline rather than the detailed step-by-step picture.

x%5E2-14x%2B3y%2B70=15%2B9y-y%5E2 can be written as %28x-7%29%5E2+%2B+%28y-3%29%5E2+=+3
We have a circle of radius sqrt%283%29 centered at (7,3)
Compare it to the template %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2

The intersection points of the secant line y = x-3 and the circle are
and
The decimal values are approximate.

Use the exact forms of A and B to compute the midpoint.

The midpoint is: (13/2, 7/2)
Let's call this point M.

C = center of the circle = (7,3)

Use the distance formula to calculate the distance from M(13/2, 7/2) to C(7, 3)
That distance is exactly %28sqrt%282%29%29%2F2
This is the green segment that tutor Edwin has marked in his diagram.

This is the adjacent leg of the right triangle BCM where angle theta is between the green line and red line of Edwin's diagram.
The red line is sqrt%283%29 units long. It's the radius of the circle. It's also the hypotenuse of the right triangle.

cos(angle) = adjacent/hypotenuse
cos(angle MCB) = MC/BC
cos(angle MCB) = sqrt(2)/2 divide over sqrt(3)
cos(angle MCB) = sqrt(6)/6
angle MCB = arccos(sqrt(6)/6)
angle MCB = 1.15026199151093 radians

Double this to get a central angle of 2*1.15026199151093 = 2.30052398302187 radians

This central angle ACB sweeps out the area of the sector shaded in green below

I used GeoGebra to make the diagram.

area of sector = (angleInRadians/2)*r^2
area of sector = (2.30052398302187/2)*(sqrt(3))^2
area of sector = 3.4507859745328

The area of that pizza slice is roughly 3.4507859745328 square units

Subtract off the area of each triangle (refer to Edwin's calculations to get the area of one triangle)
3.4507859745328 - 2*(sqrt(5)/4) = 2.3327519857829

The area of the blue shaded segment, aka minor segment, is roughly 2.3327519857829 square units

The remaining pink shaded portion is found by this formula
majorSegmentArea = AreaOfCircle - minorSegmentArea

So,
majorSegmentArea = AreaOfCircle - minorSegmentArea
majorSegmentArea = pi*r^2 - 2.3327519857829
majorSegmentArea = pi*(sqrt(3))^2 - 2.3327519857829
majorSegmentArea = 7.09202597498648


The approximate final answer is 7.09202597498648