Question

Write the following in both standard and general form

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