RADIUS AND CENTRE OF CIRCLE

If the equation of circle is {x^2} + {y^2} + 2gx + 2fy + c = 0

Then center = (–g, –f)

and radius  = \sqrt {{g^2} + {f^2} - c}

                  Or

If the equation of circle is  {x^2} + {y^2} + 2gx + 2fy + c = 0

By completing square,

\left( {{x^2} + 2gx + {g^2}} \right) + \left( {{y^2} + 2fy + {f^2}} \right) + c = {g^2} + {f^2}

                             = {\left[ {x - \left( { - g} \right)} \right]^2} + {\left[ {y - \left( { - f} \right)} \right]^2}

                             = {\left( {\sqrt {{g^2} + {f^2} - c} } \right)^2}

Example1: Find the centre and radius of the circle {x^2} + {y^2} + 6x + 8y + 9 = 0

Solution: Comparing the given equation with the standard equation, {x^2} + {y^2} + 2gx + 2fy + c = 0

\begin{array}{l}
\,\,\,\,\,\,cg\,\, = \,\,6\,\,\,\,\,\,\,\\
 \Rightarrow \,\,\,\,\,\,g\,\,\, = \,\,3
\end{array} 

\begin{array}{l}
\,\,\,\,\,2f\, = \,\,8\,\,\,\,\,\,\,\\
 \Rightarrow \,\,f\, = \,\,4
\end{array}

and c = 9

Centre \left( { - g, - f} \right) = \left( { - 3, - 4} \right)

Radius  = \sqrt {{3^2} + {4^2} - 9\,} \,\, = 4

Therefore the circle has its centre \left( { - 3, - 4} \right) and radius 4 units.

Question 1: Find the center and radius of the circle {x^2} + {y^2} = {7^2}

Solution 1: This is of the form {x^2} + {y^2} = {r^2}

Therefore the centre of the circle = \left( {0,0} \right)

Comparing the two equations radius = 7 units

Question 2: Find the centre and radius of the circle {\left( {x - 5} \right)^2} + {\left( {y + 6} \right)^2} = 81

Solution 2: This is of the form {\left( {x - h} \right)^2} + {\left( {y - k} \right)^2}\,\, = {r^2}

Comparing these two equations we get

h = 5 and y + 6and can be rewritten as y - \left( 6 \right)\, \Rightarrow \,\,\,\,\,\,k =  - 6.

 {r^2} = 81\,

Radius,r  = 9 units.

center = \left( {5, - 6} \right)

Post Author: E-Maths