Respuesta :
The circle equation in standard form: [tex](x-3)^{2}+(y+2)^{2}=4^{2}[/tex]
The circle equation in general form: [tex]x^{2}+y^{2}-6 x+4 y-3=0[/tex]
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 [tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]
where (h, k) is center and r is radius.
So the standard form of circle is
[tex]\begin{array}{l}{\left.(x-3)^{2}+(y-(-2))\right)^{2}=4^{2}} \\\\ {\rightarrow(x-3)^{2}+(y+2)^{2}=4^{2}}\end{array}[/tex]
Finding general form:
Now, we just have to expand the standard form to get the general form.
[tex]\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}[/tex]
Hence the standard and general form are found out
