Question

Find the Radius and the center of the equation, then graph the circle.

Answer 1

GIven data:

The given circle equation is,

`x^2-4x+y^3+8y=-11\ldots(1)`

To find:

(i) the centre and the radius from the given circle equation.

(ii) to graph the circle.

(i)

to find the centre and the radius using the circle equation we need the general equation and compare it with the given equation and find the centre and radius of the circle.

Because the centre of the circle in genral given by,

`C=(-g,-f)\ldots(2)`

and the radius of the circle is given by,

`r=\sqrt[]{g^2-f^2-c}\ldots(3)`

The general form of the circle equation is,

`x^2+y^2+2gx+2fy+c=0\ldots(4)`

Now we need to compare the equation with given equation, but the given the given circle equation is misarraged so , let us reorder the equation as per the general equation as follows.

`x^2+y^2-4x+8y+11=0\ldots(5)`

Now, let us compare the given equation (5) with the general equation (4) as shown bellow

`\begin{gathered} x^2=x^2 \\ y^2=y^2 \\ 2gx=-4x=2(-2)x\text{ ; g=-2} \\ 2fy=8y=2(4)y\text{ ; f=4} \\ c=11 \end{gathered}`

thus, we find the values of g, f, c, where

`\begin{gathered} g=-2 \\ f=4 \\ c=11 \end{gathered}`

Now that we know the values of g, f, c. let us subsitute in the equation (2)

to find the centre of the circle.

`\begin{gathered} C=(-g,-f) \\ g=-2;f=4 \\ C=(-(-2),-4) \\ C=(2,-4) \end{gathered}`

Thus, the centre of the circl is (2,-4).

Now, to find the radius of the circle use the equation (3) and subsitute the necessary values,

`\begin{gathered} r=\sqrt[]{g^2+f^2-c} \\ g=-2,f=4,c=11 \\ r=\sqrt[]{(-2)^2+(4)^2-11} \\ r=\sqrt[]{4+16-11} \\ r=\sqrt[]{9} \\ r=3 \end{gathered}`

Thus, the radius of the circle is r = 3.

(ii)

Now, let us graph the circle,

the radius is (2,-4) there is a negtaive in the y axis, then the point show be in the fourth quardant.

Since the radius is 3, the cricle is drawn as shown in the figure below.

The graph of the circle is,