Question

Find the centre and radius of the circle ${{x}^{2}}+{{y}^{2}}-x+2y-3=0$

Answer 1

Hint: Complete the square for ${{x}^{2}}-x$ and ${{y}^{2}}+2y$and hence express the given equation in ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{c}^{2}}$. Shift the origin to point point (a,b) and hence find the radius of the circle with respect to shifted origin. Since the radius of the circle will not change by shifting the origin, determine the radius of the circle. Use the fact that the centre of the circle ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ is (0,0) to determine the centre of the circle with respect to shifted origin and hence find the radius with respect to original coordinate system. Alternatively, use the fact that the radius of the circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is given by $r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$ and the centre of the circle is given by $\left( -g,-f \right)$. Hence determine the radius and the centre of the circle.

Complete step by step answer:

We have ${{x}^{2}}+{{y}^{2}}-x+2y-3=0$

Rewriting the above equation as follows, we get

$\left( {{x}^{2}}-x \right)+\left( {{y}^{2}}+2y \right)-3=0$

In the first term add and subtract ${{\left( \dfrac{1}{2} \right)}^{2}}$ and in the second term add and subtract 1, we get

$\left( {{x}^{2}}-x+\dfrac{1}{4}-\dfrac{1}{4} \right)+\left( {{y}^{2}}+2y+1-1 \right)-3=0$

We know that ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$

Hence, we have

${{x}^{2}}-2\times x\times \dfrac{1}{2}+{{\left( \dfrac{1}{2} \right)}^{2}}={{\left( x-\dfrac{1}{2} \right)}^{2}}$

Similarly, we know that ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$

Hence, we have

${{y}^{2}}+2\times y\times 1+{{1}^{2}}={{\left( y+1 \right)}^{2}}$

Hence, we have

$\begin{aligned}

& {{\left( x-\dfrac{1}{2} \right)}^{2}}+{{\left( y+1 \right)}^{2}}-\dfrac{1}{4}-1-3=0 \\

& \Rightarrow {{\left( x-\dfrac{1}{2} \right)}^{2}}+{{\left( y+1 \right)}^{2}}=\dfrac{17}{4} \\

\end{aligned}$

We know that if we shift the origin to point (a,b) and define the new coordinate system as (X,Y), then $X=x-a,Y=y-b$

Shifting the origin to $\left( \dfrac{1}{2},-1 \right)$, we get

$X=x-\dfrac{1}{2},Y=y+1$

Hence, we have

${{X}^{2}}+{{Y}^{2}}={{\left( \dfrac{\sqrt{17}}{2} \right)}^{2}}$

Hence according to new coordinate system, radius is $\dfrac{\sqrt{17}}{2}$ and centre is (0,0)

Since radius will not change by changing the origin, we have radius with respect to original coordinate system is $\dfrac{\sqrt{17}}{2}$

Also, we have $x=X+\dfrac{1}{2},y=Y-1$

Hence according to original coordinate system, the coordinates of the centre are $\left( 0+\dfrac{1}{2},0-1 \right)=\left( \dfrac{1}{2},-1 \right)$

Hence the centre of the circle is $\left( \dfrac{1}{2},-1 \right)$ and the radius is $\dfrac{\sqrt{17}}{2}$.

Note: Alternative Solution:

We know that the radius of the circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is given by $r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$ and the centre of the circle is given by $\left( -g,-f \right)$

Comparing the equation of the circle with the general equation of the circle, we get

$2g=-1\Rightarrow g=\dfrac{-1}{2},2f=2\Rightarrow f=1,c=-3$

Hence, we have

Coordinate of the centre are $\left( -g,-f \right)=\left( \dfrac{1}{2},-1 \right)$ and the radius of the circle is $\sqrt{{{\left( \dfrac{-1}{2} \right)}^{2}}+{{1}^{2}}-\left( -3 \right)}=\dfrac{\sqrt{17}}{2}$

Hence the centre of the circle is $\left( \dfrac{1}{2},-1 \right)$ and the radius is $\dfrac{\sqrt{17}}{2}$, which is the same as obtained above.