Question

Write the following in both standard and general form

Center: (3,-2) radius = 4
Please help!!!

Answer 1

The circle equation in standard form:
`(x-3)^(2)+(y+2)^(2)=4^(2)`

The circle equation in general form:
`x^(2)+y^(2)-6 x+4 y-3=0`

Solution:

Given that, center of a circle is (3, -2) and radius = 4.

We have to find the standard and general form of the circle.

Finding standard form:

The standard form of circle is
`(x-h)^(2)+(y-k)^(2)=r^(2)`

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

So the standard form of circle is

`\begin{array}{l}{\left.(x-3)^(2)+(y-(-2))\right)^(2)=4^(2)} \\\\ {\rightarrow(x-3)^(2)+(y+2)^(2)=4^(2)}\end{array}`

Finding general form:

Now, we just have to expand the standard form to get the general form.

`\begin{array}{l}{\text { So, }(x-3)^(2)+(y+2)^(2)=4^(2)} \\\\ {\rightarrow x^(2)+9-6 x+y^(2)+4+4 y=16} \\\\ {\rightarrow x^(2)+y^(2)-6 x+4 y+13-16=0} \\\\ {\rightarrow x^(2)+y^(2)-6 x+4 y-3=0}\end{array}`

Hence the standard and general form are found out