Question

How do you find the centre and radius of \[{x^2} + {y^2} - 6x = 0\]?

Answer 1

Hint: In this question we are asked to find the centre and radius of the given equation which we have to transform the equation to the form of the equation of circle i.e.,\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\], where \[\left( {h,k} \right)\] is the centre of the circle and\[r\]is the radius of the circle, after transforming we will get the required centre and radius of the given circle.

Complete step-by-step solution:
Given equation is \[{x^2} + {y^2} - 6x = 0\],
Now add 9 to both sides of the equation we get,
\[ \Rightarrow {x^2} + {y^2} - 6x + 9 = 9\],
Now grouping the \[x\] terms and 9 we get,
\[ \Rightarrow \left( {{x^2} - 6x + 9} \right) + {y^2} = 9\],
Now the term \[{x^2} - 6x + 9\] can be written as, \[{\left( x \right)^2} - 2\left( {3x} \right) + {\left( 3 \right)^2}\] and it is in the form \[{a^2} - 2ab + {b^2}\] which is equal to \[{\left( {a - b} \right)^2}\] where \[a = x\] and \[b = 3\],
By substituting we get \[{\left( {x - 3} \right)^2}\],
Now the equation becomes,
\[ \Rightarrow {\left( {x - 3} \right)^2} + {y^2} = 9\],
Here 9 can be written as \[{3^2}\], again the equation becomes,
\[ \Rightarrow {\left( {x - 3} \right)^2} + {\left( {y - 0} \right)^2} = {3^2}\],
Which is in form of circle \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\], where \[\left( {h,k} \right)\] is the centre of the circle and \[r\] is the radius of the circle,
Match the values in this circle to those of the standard form. The variable \[r\] represents the radius of the circle,\[h\] represents the \[x\]-offset from the origin, and \[k\] represents the \[y\]-offset from origin.
\[h = 3\],\[k = 0\] and \[r = 3\],
Now as \[\left( {h,k} \right)\] is the centre of the circle so the centre of the given circle is \[\left( {3,0} \right)\] and radius is equal to 3.
This is represented in the graph as,

\[\therefore \]The centre of the given circle \[{x^2} + {y^2} - 6x = 0\] is equal to \[\left( {3,0} \right)\] and the radius of the given circle \[{x^2} + {y^2} - 6x = 0\] is equal to 3.

Note: In solving these type of questions we should know the general equation of the circle \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] also it is important to note that we should use the complete the square method with respect to \[x\] and \[y\] so that we obtain the equation in general from. And another general equation of a circle is \[{x^2} + {y^2} + 2gx + 2fy + c = 0\].