Question

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?

Answer 1

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.