Question

How do you find the center and radius of \[{{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=4\]?

Answer 1

Hint: We know that the conditions for the equation to represent a circle is that there should be no term having \[xy\], and coefficients of \[{{x}^{2}}\And {{y}^{2}}\] should be the same. The standard form of the equation of the circle is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]. We can use the coefficients of the equation to find the center, and the radius of the circle, as follows
The center of the circle is at \[\left( -g,-f \right)\], and the radius of the circle is \[\sqrt{{{g}^{2}}+{{f}^{2}}-c}\].

Complete step by step answer:
The given equation of the circle is \[{{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=4\]. We need to convert the equation to its standard form. Expanding the bracket of the equation, we get
\[\begin{align}
  & \Rightarrow {{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=4 \\
 & \Rightarrow {{x}^{2}}-2(2)(x)+{{2}^{2}}+{{y}^{2}}+2(3)(y)+{{3}^{2}}=4 \\
 & \Rightarrow {{x}^{2}}-4x+4+{{y}^{2}}+6y+9=4 \\
 & \Rightarrow {{x}^{2}}-4x+{{y}^{2}}+6y+13=4 \\
\end{align}\]
Subtracting 4 from both sides of the above equation, we get
\[\begin{align}
  & \Rightarrow {{x}^{2}}-4x+{{y}^{2}}+6y+13-4=4-4 \\
 & \Rightarrow {{x}^{2}}-4x+{{y}^{2}}+6y+9=0 \\
\end{align}\]
The above equation can also be expressed as
\[\Rightarrow {{x}^{2}}+(2)(-2)x+{{y}^{2}}+(2)(3)y+9=0\]
The above equation is the standard form of the equation of the given circle. For a circle of standard form \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\], the center of the circle is at \[\left( -g,-f \right)\], and the radius of the circle is \[\sqrt{{{g}^{2}}+{{f}^{2}}-c}\].
Using this for the standard form of the given circle, we get
Centre of the circle at \[\left( -(-2),-3 \right)=\left( 2,-3 \right)\]. And the radius of the circle is \[\sqrt{{{(-2)}^{2}}+{{(3)}^{2}}-9}=\sqrt{4+9-9}=\sqrt{4}=2\].

Note: We can also use a different form of circle called center-radius form. This form is of the type \[{{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}\], here the center of the circle is \[(a,b)\], and the radius of the circle is \[r\]. We can convert the given equation to its center-radius form as follows,
 \[\begin{align}
  & \Rightarrow {{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=4 \\
 & \Rightarrow {{\left( x-2 \right)}^{2}}+{{\left( y-(-3) \right)}^{2}}={{2}^{2}} \\
\end{align}\]
Comparing the center-radius form, we get \[a=2,b=-3\And r=2\]. Hence, the radius of the circle is \[r=2\], and the center of the circle is at \[\left( 2,-3 \right)\].