Question

Okay so like for an equation like ${x^2} + {y^2} + 4y - 12 = 0$….how would I convert that to the standard form of a circle, ${(x - h)^2} + {(y - k)^2} = {r^2}$ ?

Answer 1

In this question, we are given an algebraic equation in terms of two unknown variable quantities “x” and “y”, both x and y have the highest exponent as 2, so the given equation is the equation of a circle. We have to convert this equation to the standard form, ${(x - h)^2} + {(y - k)^2} = {r^2}$ is the standard form of the equation of a circle, where $(h,k)$ are the coordinates of the centre of the circle and $r$ is the radius of the circle (the distance between the centre of the circle and any point on the boundary of the circle is known as the radius of the circle). We will do it by using some arithmetic identities.

Complete step-by-step solution:
We are given that ${x^2} + {y^2} + 4y - 12 = 0$
We know that the 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} + {y^2} + 4y + 4 - 4 - 12 = 0 \\\ \Rightarrow {x^2} + {y^2} + 4y + {(2)^2} - 16 = 0 \\\$
We know that ${a^2} + {b^2} + 2ab = {(a + b)^2}$ , and We can write ${x^2} = {x^2} + {(0)^2} - 2 \times 0 \times x = {(x - 0)^2}$ using these two in the obtained equation, we get –
$\Rightarrow {(x - 0)^2} + {(y + 2)^2} - 16 = 0 \\\ \Rightarrow {(x - 0)^2} + {(y + 2)^2} = {(4)^2} \\\$
Hence ${x^2} + {y^2} + 4y - 12 = 0$ in the standard form is written as ${(x - 0)^2} + {(y + 2)^2} = {(4)^2}$ .

Note: After converting the given equation into the standard way, we can also use it to draw its graph as we will get the coordinates of the centre and the radius of the circle. The coordinates of the centre of this circle is $(0, - 2)$ and the radius of this circle is 4 units. There are various ways to write the equation of a circle, but the standard way is ${(x - h)^2} + {(y - k)^2} = {r^2}$ .