Question

Write an equation in general form of the circlewith the given properties.Center at(-8, 9)and passing through the origin

Answer 1

- The general equation of a circle is given by:

`\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ \text{where, } \\ \text{The center of the circle is at point }(a,b) \\ \text{The rad}ius\text{ of the circle is }r \end{gathered}`

- Since the center of the circle is at (-8, 9), then it implies that

`\begin{gathered} a=-8 \\ b=9 \end{gathered}`

- Thus, we can write the equation as follows

`\begin{gathered} (x-(-8))^2+(y-9)^2=r^2 \\ (x+8)^2+(y-9)^2=r^2 \end{gathered}`

- As it is, the equation is still incomplete because we don't have the radius (r).

- This radius can be gotten by applying the condition given in the question that "The circle passes through the origin" .

- This means that the circle, when drawn, passes through the point (0, 0).

- Thus, we can substitute x = 0 and y = 0 into our incomplete equation and find the value of r.

- This is done below:

`\begin{gathered} (x+8)^2+(y-9)^2=r^2 \\ \\ \text{put }x=0,y=0 \\ \\ (0+8)^2+(0-9)^2=r^2 \\ 8^2+9^2=r^2 \\ r^2=64+81 \\ r^2=145 \end{gathered}`

- Thus, the equation of the circle with center (-8, 9) is:

`(x+8)^2+(y-9)^2=145`

Final Answer

The answer is

`(x+8)^2+(y-9)^2=145`