Question

How do you find the centre of a circle, given the equation \[{{x}^{2}}+{{y}^{2}}+6x-4y+3=0\] ?

Answer 1

Hint: Now we have to find the centre of the circle, whose given equation is \[{{x}^{2}}+{{y}^{2}}+6x-4y+3=0\]. We know that the general equation of the circle is \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]. We also know that the centre of the circle is \[\left( -g,-f \right)\]. By comparing the general equation of the circle and the given equation of the circle, we can get the value of g and f, which is the point at the centre of the circle.

Complete step-by-step solution:
We know that the general equation of the circle is of the form
\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]……… (1)
The given equation is of the form
\[{{x}^{2}}+{{y}^{2}}+6x-4y+3=0\]………… (2)
Now we have two equations.
We also know that the centre of the circle is \[\left( -g,-f \right)\] .

Now we can compare both the equations (1) and (2).
By comparing the coefficient of x in the general equation of the circle and the given equation, we get,
\[\begin{align}
  & \Rightarrow 2g=6 \\
 & \Rightarrow g=\dfrac{6}{2} \\
 & \Rightarrow g=3 \\
\end{align}\]
By comparing the coefficient of y in the general equation and the given equation, we get,
\[\begin{align}
  & \Rightarrow 2f=-4 \\
 & \Rightarrow f=\dfrac{-4}{2} \\
 & \Rightarrow f=-2 \\
\end{align}\]
Now we got the point \[\left( -g,-f \right)\] which is the centre of the circle.
Therefore, the centre of the circle is \[\left( -3,2 \right)\].

Note: Students may make mistakes in comparing the general equation with the given equation, as the general equation has 2g and 2f as x and y coefficient respectively, so the value of g and f is dividing the coefficient of x and y in the given equation by 2 respectively. Students may also make mistakes in symbols like plus and minus, while comparing the equation, you should also check for the correct symbol to be used.