Question 1203378
<pre>
First we get the equation x^2-14x+3y+70=15+9y-y^2 into standard form:

{{{x^2-14x+3y+70=15+9y-y^2}}}
{{{x^2-14x+y^2+3y-9y=15-70}}}
{{{x^2-14x+y^2-6y=-55}}}
completing the square:
{{{(x^2-14x+49)+(y^2-6y+9)=-55+49+9}}}
{{{(x-7)^2+(y-3)^2=3}}}
So it's a circle with center (7,3) and radius {{{sqrt(3)}}}

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

{{{x^2-14x+y^2-6y=-55}}}
{{{x^2-14x+(x-3)^2-6(x-3)=-55}}}
{{{x^2-14x+(x^2-6x+9)-6x+18=-55}}}
{{{x^2-14x+x^2-6x+9-6x+18=-55}}}
{{{2x^2-26x+27=-55}}}
{{{2x^2-26x+82=0}}}
{{{x = (-(-26) +- sqrt((-26)^2-4*2*82 ))/(2*2) }}}
{{{x = (26 +- sqrt(676-656))/(4) }}}
{{{x = (26 +- sqrt(20))/(4) }}}
{{{x = (26 +- sqrt(4*5))/(4) }}}
{{{x = (26 +- 2sqrt(5))/(4) }}}
{{{x = (2(13 +- sqrt(5)))/(4) }}}
{{{x = (13 +- sqrt(5))/(2) }}}

{{{y=x-3}}}
{{{y=(13 +- sqrt(5))/(2)-3}}} 
{{{y=(13 +- sqrt(5))/(2)-6/2}}}
{{{y=(7 +- sqrt(5))/(2)}}}
So the points of intersection are {{{(matrix(1,3,(13 +- sqrt(5))/(2),",",(7 +- sqrt(5))/(2)))}}}
Let's graph the circle and line:

{{{drawing(400,4800/13,-2,11,-6,6,graph(400,4800/13,-2,11,-6,6),circle(7,3,sqrt(3)),
line(-4,-7,14,11) )}}} 


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 

{{{drawing(400,4800/13,-2,11,-6,6,

red(line(7,3,
(13+sqrt(5))/2,
(7+sqrt(5))/2

)),

green(line(7,3,13/2,7/2)),

graph(400,4800/13,-2,11,-6,6),circle(7,3,sqrt(3)),
line(-4,-7,14,11) )}}} 

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(1/2)(x[1]+x[2])}}}{{{""=""}}}{{{expr(1/2)((13 - sqrt(5))/(2)+(13 + sqrt(5))/(2))}}}{{{""=""}}}{{{expr(1/2)(26/2)}}}{{{""=""}}}{{{13/2}}}

{{{expr(1/2)(y[1]+y[2])}}}{{{""=""}}}{{{expr(1/2)((7 - sqrt(5))/(2)+(7 + sqrt(5))/(2))}}}{{{""=""}}}{{{expr(1/2)(14/2)}}}{{{""=""}}}{{{7/2}}}

So the midpoint of the chord is {{{(matrix(1,3,13/2,",",7/2))}}}

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

{{{sqrt((7-13/2)^2+(3-7/2)^2)}}}{{{""=""}}}{{{sqrt((14/2-13/2)^2+(6/2-7/2)^2)}}}{{{""=""}}}{{{sqrt((1/2)^2+(-1/2)^2)}}}{{{""=""}}}{{{sqrt(1/4+1/4)}}}{{{""=""}}}{{{sqrt(2/4)}}}{{{""=""}}}{{{sqrt(2)/2}}}  

So the green leg of the right triangle is {{{sqrt(2)/2}}} 

We use the Pythagorean theorem to find the other leg of the right triangle,
The hypotenuse is the radius of the circle {{{sqrt(3)}}}.

other leg = half of chord = {{{sqrt(sqrt(3)^2-(sqrt(2)/2)^2)}}}{{{""=""}}}{{{sqrt(3-2/4)}}}{{{""=""}}}{{{sqrt(6/2-1/2)}}}{{{""=""}}}{{{sqrt(5/2)}}}{{{""=""}}}{{{sqrt((5*2)/(2*2))}}}{{{""=""}}}{{{sqrt(10)/2}}}

So the area of the right triangle is

{{{expr(1/2)(base)(height)}}}{{{""=""}}}{{{expr(1/2)*(sqrt(10)/2)*(sqrt(2)/2)}}}{{{sqrt(20)/8}}}{{{""=""}}}{{{sqrt(4*5)/8}}}{{{""=""}}}{{{2sqrt(5)/8}}}{{{""=""}}}{{{sqrt(5)/4}}}

We find the angle between the red and green lines from

{{{cos(angle)=(sqrt(2))/2)/(sqrt(3))=0.4082482905}}} 
{{{angle=1.150261991}}}{{{radians}}}


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

{{{drawing(400,4800/13,-2,11,-6,6,

red(line(7,3,
(13+sqrt(5))/2,
(7+sqrt(5))/2

),line(7,3,
(13-sqrt(5))/2,
(7-sqrt(5))/2

)),

green(line(7,3,13/2,7/2)),

graph(400,4800/13,-2,11,-6,6),circle(7,3,sqrt(3)),
line(-4,-7,14,11) )}}} 

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(1.150261991)}}}{{{radians}}} or {{{2.300523982}}}{{{radians}}}

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

{{{2pi-2.300523982}}}{{{radians}}}{{{""=""}}}{{{3.982661325}}}{{{radians}}}

The area of the sector = {{{expr(1/2)(radius)(matrix(1,3,angle,in,radians))}}}{{{""=""}}}

{{{expr(1/2)(sqrt(3)^"")(3.982661325)}}}{{{""=""}}}{{{3.449085882}}}

Finally we add the area of the two right triangles:

{{{3.449085882+2*(sqrt(5)/4)}}}{{{""=""}}}{{{4.567119871}}} square units

Edwin</pre>