Question

How do you find the center and radius of the circle $2{{\left( x-3 \right)}^{2}}+2{{y}^{2}}=8$?

Answer 1

First simplify the expression, if necessary to convert the expression to the standard form of circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{\left( r \right)}^{2}}$. Then find the center and the radius by comparing the obtained expression with the general equation of the circle.

Complete step by step solution:
The standard form of general equation of circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{\left( r \right)}^{2}}$ where (h, k) is the center and ‘r’ is the radius.
The expression we have $2{{\left( x-3 \right)}^{2}}+2{{y}^{2}}=8$
To convert it to standard form first we have to get rid of the coefficients of terms containing ‘x’ and ‘y’.
Dividing both sides by ‘2’, we get
$\begin{aligned} & 2{{\left( x-3 \right)}^{2}}+2{{y}^{2}}=8 \\\ & \Rightarrow \dfrac{2{{\left( x-3 \right)}^{2}}+2{{y}^{2}}}{2}=\dfrac{8}{2} \\\ & \Rightarrow {{\left( x-3 \right)}^{2}}+{{y}^{2}}=4 \\\ \end{aligned}$
It can be written as
$\Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( 2 \right)}^{2}}$
Now it is in the required form.
Comparing the above equation with the general equation of circle, we get
h=3, k=0 and r=2
Hence, the center of the given circle is (3,0) and radius is ‘2’.
This is the required solution of the given question.

Note: In the general equation of circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{\left( r \right)}^{2}}$, ‘h’ and ‘k’ respectively represents the ‘x’ and ‘y’ offset from the origin. The above circle can be graphed by taking the center as (3,0) and the radius as ‘2’.

From the graph, we can conclude that the center of the given circle lies on the ‘x’- axis at ‘3’ units towards the right of the origin.