Answer
Verified
417.6k+ views
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.
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.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
What planet is so light it could float in water class 11 physics CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Which are the Top 10 Largest Countries of the World?
Why is there a time difference of about 5 hours between class 10 social science CBSE
How do you graph the function fx 4x class 9 maths CBSE
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE