Show that, #x=1/2(3+5cos(t))#, #y=1/2(-4+5sin(t))# represent a circle passing through the origin.find co-ordinates of centre and length of radius of the circle?

1 Answer
Feb 27, 2018

Given:

#x=1/2(3+5cos(t))" [1]"#

#y=1/2(-4+5sin(t))" [2]"#

Distribute the 1/2 through the ()s for both equations:

#x=3/2+5/2cos(t)" [1.1]"#

#y=-2+5/2sin(t)" [2.1]"#

Subtract 3/2 from both sides of equation [1.1] and add 2 to both sides of equation [2.1]:

#(x-3/2)=5/2cos(t)" [1.2]"#

#(y+2)=5/2sin(t)" [2.2]"#

Square both sides of both equations:

#(x-3/2)^2=(5/2)^2cos^2(t)" [1.3]"#

#(y+2)^2=(5/2)^2sin^2(t)" [2.3]"#

Add equation [1.3] to equation [2.3]:

#(x-3/2)^2+(y+2)^2=(5/2)^2sin^2(t)+(5/2)^2cos^2(t)" [3]"#

Write #+2# as #- (-2)# and factor out #(5/2)^2# from both terms on the right:

#(x-3/2)^2+(y-(-2))^2=(5/2)^2(sin^2(t)+cos^2(t))" [3.1]"#

Use the identity #sin^2(t)+cos^2(t)= 1#:

#(x-3/2)^2+(y-(-2))^2=(5/2)^2" [3.3]"#

The general Cartesian form for circle is #(x-h)^2+(y-k)^2=r^2# where #(h,k)# is the center and r is the radius. By observation, we see that the center is:

#(3/2,-2)#

And the radius is:

#5/2#

To show that it passes through the origin, substitute 0 for x and 0 for y; the resulting equation must be true:

#(0-3/2)^2+(0-(-2))^2=(5/2)^2#

#9/4+4 = (5/2)^2#

#25/4 = (5/2)^2#

#(5/2)^2 = (5/2)^2# Q.E.D.