Question

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

Answer 1

Hint: This type of question is based on the concept of the equation of a circle. Here, to explain the concept let us consider the standard equation of a circle, that is, \[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\]. Here, (h,k) is the centre of the circle and ‘r’ is the radius. By comparing the standard equation of circle and the given circle, we get h=3 and k=0. Thus, we get the centre of the given circle. Similarly, comparing the standard equation of circle with the given circle, we get \[{{r}^{2}}=4\]. Taking the square root on both the sides, we get the radius of the circle.

Complete answer:
According to the question, we are asked to find the radius and center of the circle \[{{\left( x-3 \right)}^{2}}+{{y}^{2}}=4\].
We have been given the equation of the circle is \[{{\left( x-3 \right)}^{2}}+{{y}^{2}}=4\]. --------(1)

Let us first consider the standard equation of a circle.
\[{{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}\] ---------(2)
Here, h is a point in the x-axis and k is a point in the y-axis.
Therefore, the centre of this circle is (h,k).
Also the radius of the circle is ‘r’.
On comparing equation (1) with equation (2), we get
h=3 and k=0.
Therefore, the centre of the circle is (3,0).
Also by comparing the standard equation of circle with the given circle,
We get \[{{r}^{2}}=4\]. --------(3)
Let us take square root on both sides of the equation (3).
\[\Rightarrow \sqrt{{{r}^{2}}}=\sqrt{4}\]
But we know that \[{{2}^{2}}=4\].
\[\Rightarrow \sqrt{{{r}^{2}}}=\sqrt{{{2}^{2}}}\]
Also we know that \[\sqrt{{{x}^{2}}}=x\].
Using this in the above obtained expression, we get
\[r=2\]
Therefore, the radius of the circle is 2 units.
Hence, the centre and radius of the circle \[{{\left( x-3 \right)}^{2}}+{{y}^{2}}=4\] is (3,0) and 2 respectively.

Note: We can also solve this type of question by plotting the graph.
The given equation is \[{{\left( x-3 \right)}^{2}}+{{y}^{2}}=4\].
\[\Rightarrow {{y}^{2}}=4-{{\left( x-3 \right)}^{2}}\]
On taking square root on both the sides, we get,
\[\Rightarrow \sqrt{{{y}^{2}}}=\sqrt{4-{{\left( x-3 \right)}^{2}}}\]
\[\Rightarrow y=\sqrt{4-{{\left( x-3 \right)}^{2}}}\]
On substituting any value of x from x- axis, we get a corresponding term y from y-axis.
On plotting a graph with the obtained points, we get

Therefore, from the circle obtained from plotting the graph, we find that the centre is (3,0) and radius is 2 units.