Question

How do you find the equation of a circle center of $(5,-2)$ and a radius of $2$?

Answer 1

Step-1: Discuss the equation of a circle

Center of the circle $=(5,-2)$ and radius $=2\mathrm{units}.$

The general equation of a circle is given by ${(\mathrm{x}-\mathrm{\alpha })}^2+{(\mathrm{y}-\mathrm{\beta })}^2={\mathrm{r}}^2---\left(1\right)$ a center of $(0,0)$ and a radius of $r$.

Where $(\mathrm{x},\mathrm{y})$ is any point on the given circle, $(\mathrm{\alpha },\mathrm{\beta })$ is the center of the given circle and $\mathrm{r}$ is the radius of the given circle.

Step-2: Determine the equation of the circle:

Put the values of the center and the radius in the equation (1) :

Evaluate the expression :

$\begin{array}{r}\Rightarrow (x-(5){)}^2+(y-(-2){)}^2=2^2\\ \Rightarrow (x-5{)}^2+(y+2{)}^2=2^2\end{array}$

Solve the equation using the following identities :

$\begin{array}{rcl}{(\mathrm{x}+\mathrm{y})}^2& =& {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{xy}\\ {(\mathrm{x}-\mathrm{y})}^2& =& {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{xy}\end{array}$

Evaluate the expression further :

$$\begin{gathered} \Rightarrow x^2+5^2-2\cdot 5\cdot x+y^2+2^2+2\cdot 2\cdot y=4 \\ \Rightarrow x^2+25-10x+y^2+4+4y=4 \\ \begin{array}{rl}& \Rightarrow x^2+y^2-10x+4y=4-25-4\\ & \Rightarrow x^2+y^2-10x+4y=-25\end{array} \\ \Rightarrow x^2+y^2-10x+4y+25=0 \end{gathered}$$

Hence, a circle having its center as $(5,-2)$ and radius $=2$ units will have its equation as

$\Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-10\mathrm{x}+4\mathrm{y}+25=0$