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 the radius is 2 units.