Question

How do you write ${x^2} + {y^2} - x + 2y + 1 = 0$ into standard form?

Answer 1

Hint: We know that we use algebra in one of the most important branches of mathematics; the given equation is a combination of numerical values and alphabets so it is an algebraic expression. The unknown variables x and y both are raised to the power 2, so the given equation is definitely not a polynomial equation, it is an equation of a circle. There are various ways to write the equation of a circle, but the standard way is ${(x - h)^2} + {(y - k)^2} = {r^2}$ . so we will convert the given equation into standard form with the help of arithmetic identities.

Complete step-by-step solution:
We are given that ${x^2} + {y^2} - x + 2y + 1 = 0$
We know that standard form of the equation of a circle is ${(x - h)^2} + {(y - k)^2} = {r^2}$
So we will add and subtract some terms in the given question to convert them into standard form –
$
  {x^2} - x + \dfrac{1}{4} - \dfrac{1}{4} + {y^2} + 2y + 1 - 1 + 1 = 0 \\
   \Rightarrow {x^2} - 2 \times \dfrac{1}{2} \times x + {(\dfrac{1}{2})^2} - \dfrac{1}{4} + {y^2} + 2 \times 1 \times y + {(1)^2} - 1 + 1 = 0 \\
 $
We know that ${a^2} + {b^2} - 2ab = {(a - b)^2}$ and ${a^2} + {b^2} + 2ab = {(a + b)^2}$ , using these two identities in the obtained equation, we get –
Hence, the standard form of ${x^2} + {y^2} - x + 2y + 1 = 0$ is ${(x - \dfrac{1}{2})^2} + {(y + 1)^2} = \dfrac{1}{4}$ .

Note: The obtained equation is an equation of the circle. To plot a circle on the graph paper, we use its equation. The standard form of the equation of a circle is ${(x - h)^2} + {(y - k)^2} = {r^2}$ where $(h,k)$ are the coordinates of the centre of the circle and $r$ is the radius of the circle (radius is defined as the distance between the centre of the circle and any point on the boundary of the circle). Thus the coordinates of the centre of the given circle are $(\dfrac{1}{2}, - 1)$ and its radius is equal to $\dfrac{1}{2}units$.