
What is the equation of the circle with centre at \[\left( {0,0} \right)\] and radius of 7?
Answer
521.4k+ views
Hint: Here in this question, we have to find the equation of a circle using a given point and radius. The standard equation for a circle, which is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] , where \[\left( {h,k} \right)\] is the centre of the circle and \[r\] is the radius. On substituting centre and radius and by further simplification we get the required solution.
Complete step by step solution:
Equation of a Circle When the Centre is Origin: Consider an arbitrary point \[P\left( {x,y} \right)\] on the circle. Let ‘ \[r\] ’ be the radius of the circle and origin \[\left( {0,0} \right)\] can be found using the distance formula which is equal to- \[{x^2} + {y^2} = {r^2}\]
Equation of a Circle When the Centre is not an Origin: Let \[C\left( {h,k} \right)\] be the centre of the circle and \[P\left( {x,y} \right)\] be any point on the circle, the radius of a circle is ‘ \[r\] ’.
the equation of the circle with centre \[\left( {h,k} \right)\] and the radius ‘ \[r\] ’ is:
\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
which is called the standard form for the equation of a circle.
Consider, the given data in question
Centre of the circle is: \[C\left( {0,0} \right)\]
Radius of the circle: \[r = 7\]
Now, consider the standard form for the equation of a circle is: \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
Here, \[h = 0\] , \[k = 0\] and \[r = 7\] . On substituting in standard equation, we have
\[ \Rightarrow \,\,{\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {\left( 7 \right)^2}\]
On simplification, we get
\[ \Rightarrow \,\,{x^2} + {y^2} = 49\]
Hence, the equation of the circle is \[{x^2} + {y^2} = 49\] .
So, the correct answer is “ \[{x^2} + {y^2} = 49\] ”.
Note: A circle is formed when an arc is drawn from the fixed point called the centre, in which all the points on the curve are having the same distance from the centre point of the centre. We know that there is a question that arises in case of a circle whether being a function or not. It is clear that a circle is not a function.
Complete step by step solution:
Equation of a Circle When the Centre is Origin: Consider an arbitrary point \[P\left( {x,y} \right)\] on the circle. Let ‘ \[r\] ’ be the radius of the circle and origin \[\left( {0,0} \right)\] can be found using the distance formula which is equal to- \[{x^2} + {y^2} = {r^2}\]
Equation of a Circle When the Centre is not an Origin: Let \[C\left( {h,k} \right)\] be the centre of the circle and \[P\left( {x,y} \right)\] be any point on the circle, the radius of a circle is ‘ \[r\] ’.
the equation of the circle with centre \[\left( {h,k} \right)\] and the radius ‘ \[r\] ’ is:
\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
which is called the standard form for the equation of a circle.
Consider, the given data in question
Centre of the circle is: \[C\left( {0,0} \right)\]
Radius of the circle: \[r = 7\]
Now, consider the standard form for the equation of a circle is: \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
Here, \[h = 0\] , \[k = 0\] and \[r = 7\] . On substituting in standard equation, we have
\[ \Rightarrow \,\,{\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {\left( 7 \right)^2}\]
On simplification, we get
\[ \Rightarrow \,\,{x^2} + {y^2} = 49\]
Hence, the equation of the circle is \[{x^2} + {y^2} = 49\] .
So, the correct answer is “ \[{x^2} + {y^2} = 49\] ”.
Note: A circle is formed when an arc is drawn from the fixed point called the centre, in which all the points on the curve are having the same distance from the centre point of the centre. We know that there is a question that arises in case of a circle whether being a function or not. It is clear that a circle is not a function.
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