Question

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

Answer 1

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

& \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{aligned}$$ Subtracting 4 from both sides of the above equation, we get $$\begin{aligned} & \Rightarrow {{x}^{2}}-4x+{{y}^{2}}+6y+13-4=4-4 \\\ & \Rightarrow {{x}^{2}}-4x+{{y}^{2}}+6y+9=0 \\\ \end{aligned}$$ The above equation can also be expressed as $$\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{aligned} & \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{aligned}$$ Comparing the center-radius form, we get $$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)$$.