Question

what are the coordinates for the center of the circle in the length of the radius X squared plus Y squared minus X minus 2Y -11 over four equals zero

Answer 1

Given:

The equation of a circle is:

$x^2+y^2-x-2y-\dfrac{11}{4}=0$

To find:

The center of the circle and its radius.

Solution:

The standard form of a circle is:

$(x-h)^2+(y-k)^2=r^2$          ...(i)

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

The given equation of the circle is:

$x^2+y^2-x-2y-\dfrac{11}{4}=0$

Write the given equation in standard form of a circle.

$(x^2-x)+(y^2-2y)-\dfrac{11}{4}=0$

$\left(x^2-x+\left(\dfrac{1}{2}\right)^2\right)+(y^2-2y+1^2)-\dfrac{11}{4}-\left(\dfrac{1}{2}\right)^2+1^2=0$

$\left(x-\dfrac{1}{2}\right)^2+(y-1)^2-\dfrac{11}{4}-\dfrac{1}{4}-1=0$

$\left(x-0.5\right)^2+(y-1)^2-\dfrac{12}{4}-1=0$

On further simplification, we get

$\left(x-0.5\right)^2+(y-1)^2-3-1=0$

$\left(x-0.5\right)^2+(y-1)^2-4=0$

$\left(x-0.5\right)^2+(y-1)^2=4$

$\left(x-0.5\right)^2+(y-1)^2=2^2$            ...(ii)

On comparing (i) and (ii), we get

$h=0.5,k=1,r=2$

Therefore, the center of the circle is (0.5,1) and radius is 2 units.