Find a parametrization of the circle of radius 3 in the xy-plane, centered at (−2,2), oriented counterclockwise. The point (1,2) should correspond to t=0. Use t as the parameter for all of your answers.

x = 3*cos(t) - 2  

y = 3*sin(t) + 2  

Step-by-step explanation:

The equation for a circle of radius 3 and centered at the point ( h , k )           (  -2 , 2 ) is:

( x - h )²  +  (  y  - k )²  = r²

In this particular case

h = - 2   k = 2    and  r = 3

Then to parametrizace the equation

x  = r* cos(t) +  h

y = r*sin(t)  + k

x =  3*cos(t) + (- 2)       ⇒   x = 3*cos(t) - 2    

y = 3 *sin(t) + (2)          ⇒   y = 3*sin(t) + 2

Now :

when t = 0   x should be  1

Then   x = 3 cos(0) - 2      x  = 3 - 2        x  = 1

and t = 0   y should be 2

Then   y = 3*sin(t) + 2    ⇒  y =  3*sin(0) + 2       y  =  2

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐