Question

How do you find the centre and radius of the circle ${{x}^{2}}+{{y}^{2}}-6x-4y-12=0$?

Answer 1

We start solving the problem by first adding and subtracting the number 6 from the given equation of circle. We then add and subtract the number 4 from the given equation of circle. We then make the necessary arrangements in the obtained equation to convert the given equation of circle to the form resembling ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$. We then make use of the fact that the centre and radius of that circle of the form ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ are $\left( a,b \right)$ and r units to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the centre and radius of the circle ${{x}^{2}}+{{y}^{2}}-6x-4y-12=0$.
We have given the equation of the circle as ${{x}^{2}}+{{y}^{2}}-6x-4y-12=0$.
Now, let us convert this equation into the form ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$.
So, we have ${{x}^{2}}+{{y}^{2}}-6x-4y-12=0$ ---(1).
Let us both add and subtract 9 in equation (1).
$\Rightarrow {{x}^{2}}-6x+9+{{y}^{2}}-4y-12-9=0$.
$\Rightarrow {{\left( x-3 \right)}^{2}}+{{y}^{2}}-4y-21=0$ ---(2).
Let us both add and subtract 4 in equation (2).
$\Rightarrow {{\left( x-3 \right)}^{2}}+{{y}^{2}}-4y+4-21-4=0$.
$\Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}-25=0$.
$\Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=25$.
$\Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}={{5}^{2}}$ ---(3).
We know that the if a circle equation is in the form of ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$, then the centre and radius of that circle is $\left( a,b \right)$ and r units. Let us use this result in equation (3).
So, we get the centre and radius of the circle ${{\left( x-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}={{5}^{2}}$ as $\left( 3,2 \right)$ and 5 units.

$\therefore$ The centre and radius of the circle ${{x}^{2}}+{{y}^{2}}-6x-4y-12=0$ are $\left( 3,2 \right)$ and 5 units.

Note: We can also solve this problem as shown below:
We have given the equation of the circle as ${{x}^{2}}+{{y}^{2}}-6x-4y-12=0$. Let us compare this with the standard equation of circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$.
We know that the centre and radius of the circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is defined as $\left( -g,-f \right)$ and $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$.
So, the centre of the circle is $\left( \dfrac{-\left( -6 \right)}{2},\dfrac{-\left( -4 \right)}{2} \right)=\left( 3,2 \right)$ and radius as $\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -2 \right)}^{2}}-\left( -12 \right)}=\sqrt{9+4+12}=\sqrt{25}=5$ units.