Question

How do you find the center and radius for \[{x^2} + {y^2} = 36\] ?

Answer 1

Hint: Compare the given equation to the standard equation of a circle and identify the center and the radius.
The standard equation of a circle is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] where point \[\left( {h,k} \right)\] is the center and \[r\] is the radius of the circle such that \[r > 0\].

Complete step-by-step solution:
The given equation of the circle is \[{x^2} + {y^2} = 36\].
We can convert the given equation into the standard form of a circle as shown below.
\[ \Rightarrow {\left( {x - 0} \right)^2} + {\left( {y - 0} \right)^2} = {\left( 6 \right)^2}\]
Now, compare the obtained equation with standard equation of a circle \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] where point \[\left( {h,k} \right)\] is the center and \[r\] is the radius of the circle such that \[r > 0\].
It is observed that the value for \[h\] and \[k\] is \[0\] and the value for \[r\] is \[6\] as \[r < 0\] is not allowed.
Therefore, center of the circle is \[\left( {h,k} \right) = \left( {0,0} \right)\] and radius of the circle is \[r = 6\] for the given circle equation \[{x^2} + {y^2} = 36\].
Circle is a close figure, uniquely defined by the position of a fixed point (center) and the constant distance between the fixed point and the point on the circle (radius).
All the possible circles in a plane are similar.

Note: Always convert the given general equation to standard equation then compare to obtain the center and radius of the circle. If we assume the position of a center is point \[\left( {h,k} \right)\], any point on circle is \[\left( {x,y} \right)\] and the constant distance between center \[\left( {h,k} \right)\] and any point on circle \[\left( {x,y} \right)\] is radius \[r\] then according to distance formula \[\sqrt {{{\left( {x - h} \right)}^2} + {{\left( {y - k} \right)}^2}} = r\] which is equivalent to the equation of a circle in standard form \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\].