Question

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

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, $${(x-h)}^2+{(y-k)}^2=r^2$$. 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 $${(x-3)}^2+y^2=4$$.
We have been given the equation of the circle is $${(x-3)}^2+y^2=4$$. --------(1)

Let us first consider the standard equation of a circle.
$${(x-h)}^2+{(y-k)}^2=r^2$$ ---------(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 $${(x-3)}^2+y^2=4$$ is (3,0) and 2 respectively.

Note: We can also solve this type of question by plotting the graph.
The given equation is $${(x-3)}^2+y^2=4$$.
$$\Rightarrow y^2=4-{(x-3)}^2$$
On taking square root on both the sides, we get,
$$\Rightarrow \sqrt{y^2}=\sqrt{4-{(x-3)}^2}$$
$$\Rightarrow y=\sqrt{4-{(x-3)}^2}$$
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.