Question

General second-degree equation in x and y is $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ , where a, h, b, g, f, and c are constant.
Prove that conditions for it to be a circle is $a = b$ and $h = 0$.

Answer 1

Hint: At first we will take a general equation of the circle i.e. ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$ , with center $(a,b)$ and radius ‘r’. Now, we will simplify this equation to compare it with the given second-degree equation.
On comparing, we get the results that if $a = b$ and $h = 0$ , then the second-degree equation $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ forms a circle.

Complete step-by-step answer:
Given data: Second-degree equation i.e. $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$
We know that the general equation of a circle with center $(a,b)$ and radius ‘r’ is given by
 $ \Rightarrow {\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$
Using ${\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy$ , we get,
 $ \Rightarrow {x^2} + {a^2} - 2xa + {y^2} + {b^2} - 2yb = {r^2}$
On simplifying we get,
 $ \Rightarrow {x^2} + {y^2} - 2xa - 2yb + {a^2} + {b^2} - {r^2} = 0$
On comparing this equation with a second-degree equation i.e. $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$
We can say that $a = b$ and $h = 0$
Hence we proved that if $a = b$ and $h = 0$ , then the second-degree equation $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ forms a circle.

Note: An alternate method for the above solution can be
Solving for the second-degree equation i.e. $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$
 $ \Rightarrow a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$
Now for $a = b$ and $h = 0$
 $ \Rightarrow a{x^2} + a{y^2} + 2gx + 2fy + c = 0$
Dividing the whole equation with ‘a’
 $ \Rightarrow {x^2} + {y^2} + 2\dfrac{g}{a}x + 2\dfrac{f}{a}y + \dfrac{c}{a} = 0$
Now adding and subtracting ${\left( {\dfrac{g}{a}} \right)^2}$ and ${\left( {\dfrac{f}{a}} \right)^2}$ on the left-hand side of the equation
 \[ \Rightarrow {x^2} + 2\dfrac{g}{a}x + {\left( {\dfrac{g}{a}} \right)^2} + {y^2} + 2\dfrac{f}{a}y + {\left( {\dfrac{f}{a}} \right)^2} + \dfrac{c}{a} - {\left( {\dfrac{g}{a}} \right)^2} - {\left( {\dfrac{f}{a}} \right)^2} = 0\]
Now using ${x^2} + {y^2} + 2xy = {\left( {x + y} \right)^2}$
 \[ \Rightarrow {\left( {x + \dfrac{g}{a}} \right)^2} + {\left( {y + \dfrac{f}{a}} \right)^2} + \dfrac{c}{a} - {\left( {\dfrac{g}{a}} \right)^2} - {\left( {\dfrac{f}{a}} \right)^2} = 0\]
On taking the constant term to the right-hand side
 \[ \Rightarrow {\left( {x + \dfrac{g}{a}} \right)^2} + {\left( {y + \dfrac{f}{a}} \right)^2} = {\left( {\dfrac{g}{a}} \right)^2} + {\left( {\dfrac{f}{a}} \right)^2} - \dfrac{c}{a}\]
Now using $\sqrt {{x^2}} = x$
 \[ \Rightarrow {\left( {x + \dfrac{g}{a}} \right)^2} + {\left( {y + \dfrac{f}{a}} \right)^2} = {\left( {\sqrt {{{\left( {\dfrac{g}{a}} \right)}^2} + {{\left( {\dfrac{f}{a}} \right)}^2} - \dfrac{c}{a}} } \right)^2}\]
Since it is an equation of a circle with center \[\left( { - \dfrac{g}{a}, - \dfrac{f}{a}} \right)\] and radius \[\sqrt {{{\left( {\dfrac{g}{a}} \right)}^2} + {{\left( {\dfrac{f}{a}} \right)}^2} - \dfrac{c}{a}} \]
Hence we proved that if $a = b$ and $h = 0$ , then the second-degree equation $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ forms a circle