For each of the described curves, decide if the curve would be more easily given by a polar equation or a Cartesian equation. Then write an equation for the curve. a. A circle with radius 3 and center (1, 1).
b. A circle centered at the origin with radius 1.

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

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0