Answer:
a) x²+y²-2x-2y-7= 0
b) x²+y²-1 = 0
Step-by-step explanation:
Given the equation of a curve;
A circle with radius 3 and center (1, 1).
The equation of a circle is expressed as;
(x-a)² +(y-b)² = r² where;
(a, b) is the center of the circle
r is the radius of the circle
Substitute the given center and radius into the given formula;
(x-1)² +(y-1)² = 3²
(x-1)² +(y-1)² = 9
Open the bracket
x²-2x+1 + y²-2y + 1 = 9
Collect the like terms;
x²+y²-2x-2y+2 = 9
x²+y²-2x-2y+2- 9 = 0
x²+y²-2x-2y-7= 0
Hence the equation of the curve is x²+y²-2x-2y-7= 0
For a circle centered at the origin with radius 1, the equation is expressed as;
(x- 0)² + (y-0)² = 1²
x²+y² = 1
x²+y² - 1 = 0
Hence the required equation of the curve is x²+y² - 1 = 0
This shows that the curve will be more easily given by a cartesian equation
