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heureka

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 #2
avatar+26399 
+2

The general form of the equation of a circle is x2+y2−4x−8y−5=0.

What are the coordinates of the center of the circle?

 

A circle can be defined as the locus of all points that satisfy the equation

(xh)2+(yk)2=r2  ( Standard Form )

where r is the radius of the circle,
and h,k are the coordinates of its center.

 

The general Form is:

x2+y2+ax+by+c=0

 

Standard Form to general Form:

(xh)2+(yk)2=r2x22xh+h2+y22yk+k2=r2x2+y2+x(2h)=a+y(2k)=b+h2+k2r2=c=0

 

h,k and r ?

x2+y2+x(2h)=a+y(2k)=b+h2+k2r2=c=0a=2hh=a2b=2kk=b2c=h2+k2r2c=(a2)2+(b2)2r2c=a2+b24r2r2=a2+b24cr=a2+b24c

 

If we have a,b and c, we can calculate h,k and r:

x2+y2+ax+by+c=0h=a2k=b2r=a2+b24c


a=4b=8c=5

x2+y24x8y5=0h=42h=2k=82k=4r=(4)2+(8)24(5)r=16+644+5r=20+5r=25r=5

 

The coordinates of the center of the circle is (2,4) and the radius is 5

 

laugh

16.05.2017
 #7
avatar+26399 
+1

Given:

sinα+sinβ=1, cosα+cosβ=1

Find the value of

sinαcosβ

 

(1)sin(α)+sin(β)=1(2)cos(α)+cos(β)=1(1)(2):sin(α)+sin(β)(cos(α)+cos(β))=0sin(α)+sin(β)cos(α)cos(β)=0sin(α)cos(α)=2sin(α45)=cos(β)sin(β)=2sin(β45)2sin(α45)=2sin(β45)sin(α45)=sin(β45)sin(α45)=sin((β45))sin(α45)=sin(45β)α45=45βα=90β

 

sin(α)cos(β)=sin(90β)cos(β)=cos(β)cos(β)=0

 

Proof:

sin(α)+sin(β)=1|α=90βsin(90β)+sin(β)=1cos(β)+sin(β)=1|cos(β)+sin(β)=2sin(β+45)2sin(β+45)=1sin(β+45)=12β+45=arcsin(12)+n360nZβ+45=45+n360nZβ=90+n360nZα=90βα=90(90+n360)nZα=0+n360nZ

 

 

laugh

15.05.2017