Find the perimeter of the circle with the equation
\[ 4x^2 + 4y^2 - 12x + 5 = 0 \quad ① \]
Steps in solving perimeter of circle with equation.
- Recognize a circle from its equation.
- Rewrite it into standard form.
- Find its centre and radius.
- Use the radius to compute the circle's perimeter.
NB: If you have a general second-degree equation like this
\[ Ax^2 + Ay^2 + Dx + Ey + F = 0 \quad ② \]
and
The coefficient of \( x^2 \) and \( y^2 \) are the same (which is equal to \( A \)) in eqn ② and (both equal to 4) in eqn ①
If the coefficient of \( xy \) term (if any) is zero (i.e. no \( xy \) term), then the equation represents a circle)
It can always be written into the standard form
\[ (x - h)^2 + (y - k)^2 = r^2 \quad ③ \]
The fact that the coefficient of \( x^2 \) and \( y^2 \) are the same keeps the shape round and symmetric in all directions.
If the coefficients are different i.e \( Ax^2 + By^2 + \ldots = 0 \) where \( A \neq B \) then the shape is usually an ellipse (stretched circle).
Now let's follow the steps listed
Step 1. Recognize the Circle. which we've done.
i.e the coefficient of \( x^2 \) and \( y^2 \) are both 4.
no \( xy \) term, only an x-term (\(-12x\)). So this is indeed a circle.
Rewrite in standard form.
\[ 4x^2 + 4y^2 - 12x + 5 = 0 \]
\[ \Rightarrow 4x^2 - 12x + 4y^2 + 5 = 0 \]
\[ 4(x^2 - 3x) + 4y^2 + 5 = 0 \]
Divide through by 4.
\[ \frac{4(x^2-3x)}{4} + \frac{4y^2}{4} + \frac{5}{4} = 0 \]
\[ x^2 - 3x + y^2 = -\frac{5}{4} \quad ④ \]
Complete the square for the x-terms.
- identify the coefficient of \( x \) which is \(-3\).
- take half of this coefficient and square it
\[ \left(-\frac{3}{2}\right)^2 = \frac{9}{4}. \]
- add and subtract from both side of eqn ④
\[ x^2 - 3x + \frac{9}{4} + y^2 = -\frac{5}{4} + \frac{9}{4} \]
\[ x^2 - 3x + \frac{9}{4} + y^2 = 1 \Rightarrow (x - \frac{3}{2})^2 + y^2 = 1. \]
So the equation becomes
\[ (x - \frac{3}{2})^2 + y^2 = 1 \quad ⑤ \]
Equation ⑤ is in the form of eqn ③ which is \( (x-h)^2 + (y-k)^2 = r^2 \)
where Centre is \( (h,k) \) and radius is \( r \).
Compare eqn ⑤ and ③
Centre = \( -(-3/2), 0 \)
Centre = \( (\frac{3}{2}, 0) \), radius \( r^2 = 1 \).
\[ r^2 = 1 \]
square root both side \( \sqrt{r^2} = \sqrt{1} \Rightarrow r = 1 \).
Step 4 - find the perimeter (Circumference).
\[ C = 2\pi r \]
\[ = 2 \times \pi \times 1 \]
\[ = 2\pi. \]
