Question

Find the center and the radius of the circle with the equation: x^2-2x+y^2+4y+1=0?

Answer 1

Answer:

Center (h,k) is (1,-2) and radius r = 2

Step-by-step explanation:

We need to find the center and radius of the circle of the given equation:

$x^2-2x+y^2+4y+1=0$

We need to transform the above equation into standard form of circle

$(x-h)^2 + (y-k)^2 = r^2$

where(h,k) is the center of circle and r is radius of circle.

Solving the given equation:

$x^2-2x+y^2+4y+1=0$

Moving 1 to right side

$x^2-2x+y^2+4y=-1$

Now making perfect square of x^2-2x and y^2+4y

Adding +1 and +4 on both sides of the equation

$x^2-2x+1+y^2+4y+4=-1+1+4$

Now, x^2-2x+1 is equal to (x-1)^2 and y^2+4y+2 =(y+2)^2

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

Comparing with standard equation of circle:

$(x-h)^2 + (y-k)^2 = r^2$

h = 1 , k =-2 and r =2 because r^2 =4 then r=2

So, center (h,k) is (1,-2) and radius r = 2