Change r=6cos(theta)+7sin(theta) to rectangular form ?
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#(x-3)^2+(y-7/2)^2= 85/4#
#r^2=x^2+y^2#
#x=rcostheta#
#y=rsintheta#
#r^2= 6rcostheta+7rsintheta#
#x^2+y^2= 6x+7y#
#x^2-6x+9+y^2-7y+49/4=9+49/4#
#(x-3)^2+(y-7/2)^2= 85/4#
#(x-3)^2+(y-7/2)^2= (sqrt85/2)^2#
Given: #r=6cos(theta)+7sin(theta)#
Multiply both sides of the equation by #r#:
#r^2=6rcos(theta)+7rsin(theta)#
Substitute #r^2 = x^2+y^2#, #y =rsin(theta)#, and #x = rcos(theta)#:
#x^2+y^2=6x+7y#
Technically, we are done but we recognize that the equation is not in a standard form, therefore, we shall proceed.
Move everything to the left so that the equation equals 0:
#x^2-6x+y^2-7y=0#
Add #h^2+k^2# to both sides so that we can complete the squares:
#x^2-6x+ h^2+y^2-7y+k^2=h^2+k^2#
Use the middle terms to find the values of #h# and #k#:
#-2hx= -6x# and #-2ky = -7y#
#h= 3# and #k = 7/2#
Write the left side as squares and the right side as #3^2+(7/2)^2#:
#(x-3)^2+(y-7/2)^2= 3^2+(7/2)^2#
Simplify the right side:
#(x-3)^2+(y-7/2)^2= 85/4#
To comply with the standard Cartesian form of the equation of a circle, we should write the right side as a square:
#(x-3)^2+(y-7/2)^2= (sqrt85/2)^2#
Rectangular form is # (x -3)^2 +(y-3.5)^2 =21.25#
We know ,#r^2=x^2+y^2 , x= r cos theta , y= r sin theta#
# r = 6 cos theta +7 sin theta # or
# r*r = (6 cos theta +7 sin theta)*r # or
# r^2 = 6 r cos theta +7 r sin theta # or
# x^2+y^2 = 6 x +7 y# or
# x^2 -6 x +y^2 -7 y =0# or
# x^2 -6 x +9 +y^2 -7 y +3.5^2=9 +12.25# or
# (x -3)^2 +(y-3.5)^2 =21.25#
Rectangular form is # (x -3)^2 +(y-3.5)^2 =21.25#
graph{x^2+y^2= 6 x+7 y [-20, 20, -10, 10]} [Ans]
