Question

Find the area of the region enclosed by the graph of $x^2 + y^2 = 2x - 6y + 6$.

Answer 1

Complete the squares to get

$x^2+y^2=2x-6y+6$

$\implies(x^2-2x+1)+(y^2+6y+9)=16$

$\implies(x-1)^2+(y+3)^2=4^2$

which is the equation of a circle centered at (1, -3) with radius 4, and thus with area 16π.

Answer 2

The area of the region enclosed in the graph x²+y² =2x-6y+6 is 50.26 square units.

What is the area of the circle?

The area of the circle can be calculated by the product of π times square of the radius or the product of π/4 times square of diameter.

Area of the circle= A= πr²= πd²/4

where r is radius of the circle and d is diameter of the circle.

Here given that

the equation of circle is given by

x²+y² =2x-6y+6

⇒ x²+y² -2x-6y = 6

⇒ x²-2x+y²-6y = 6

Adding 1 on both sides

⇒ x²-2x+1+y²-6y = 6+1

Adding 9 on both sides

⇒ x²-2x+1+ y²-6y+9 = 6+1+9

⇒x²-2x+1+ y²-6y+9 = 16

⇒ (x-1)² + (y-3)²= 4²

which is similar to equation of the circle

(x-a)² + (y-b)²= r²

This is the equation of the circle in center and radius form where (a,b) is the center of the circle and r is the radius of the circle.

here in the equation of the circle the center of the circle is (1,3)

radius is 4 units.

then the area of the circle is = πr²= π4²= 16π= 50.26 square units.

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