Question

How do you write the equation of the circle in standard form ${{x}^{2}}+{{y}^{2}}-6x+4y-3=0$ ?

Answer 1

Hint: We are given an equation of a circle as ${{x}^{2}}+{{y}^{2}}-6x+4y-3=0$
We have to find the equation of the circle in standard form. A standard form is given as ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ where $\left( h,k \right)$ is centre and ‘r’ is radius. We will use completing the square method to convert ${{x}^{2}}+{{y}^{2}}-6x+4y-3=0$ into standard form. We will complete the square and the equation will be transformed in standard form.


Complete step-by-step solution:
We are given an equation of the circle as ${{x}^{2}}+{{y}^{2}}-6x+4y-3=0$.
We have to convert this into standard form
Standard form of the circle is given as –
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ where $\left( h,k \right)$ is centre of circle.
We will use completing the square method to change into ${{\left( x-h \right)}^{2}}$ and ${{\left( y-k \right)}^{2}}$ so that given equation ${{x}^{2}}+{{y}^{2}}-6x+4y-3=0$ transform to standard form.
Now, we have –
${{x}^{2}}+{{y}^{2}}-6x+4y-3=0$
Adding ‘3’ on both side and simplifying we get –
${{x}^{2}}+{{y}^{2}}-6x+4y=3$
Now in ${{x}^{2}}-6x$ to make a complete square we need to find the perfect number which when added will complete our square
Similarly of ${{y}^{2}}+4y$
So, ${{x}^{2}}-2\times 3x+{{y}^{2}}+2\times 2\times y=3$
So we add ${{3}^{2}}$ and also we add ${{2}^{2}}$ on both side so we get –
${{x}^{2}}-2\times 3\times x+{{3}^{2}}+{{y}^{2}}+2\times 2\times y+{{2}^{2}}=3+{{3}^{2}}+{{2}^{2}}$
Now as ${{x}^{2}}+{{3}^{2}}-2\times 3x={{\left( x-3 \right)}^{2}}$ and ${{y}^{2}}+{{2}^{2}}+2\times 2\times y={{\left( y+2 \right)}^{2}}$
so, we get –
${{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=3+9+4\text{ }\left[ \text{as }{{3}^{2}}=9\text{ and }4={{2}^{2}} \right]$
Simplifying, we get –
${{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=16$
Now as $16={{4}^{2}}$ and $y+2=\left( y-\left( -2 \right) \right)$ so we get –
${{\left( x-3 \right)}^{2}}+{{\left( y-\left( -2 \right) \right)}^{2}}={{4}^{2}}$
So we get our standard equation of circle as ${{\left( x-3 \right)}^{2}}+{{\left( y-\left( -2 \right) \right)}^{2}}={{4}^{2}}$ .

Note: While solving the square we can add things to both sides as it won't affect the equation, if we just add it to one side then the equation we get will be incorrect. Also $y-\left( -2 \right)=y+2$ because we know that the product of ‘2 negatives’ is positive.