Circle

If the equation $\;$ $3x^2 + \left(3 - p\right)xy + qy^2 - 2px = 8pq$ $\;$ represents a circle, find $p$ and $q$. Also determine the center and the radius of the circle.

The given equation is: $\hspace{1em}$ $3x^2 + \left(3 - p\right)xy + qy^2 - 2px = 8pq$ $\;\;\; \cdots \; (1)$

Standard equation of a circle is: $\hspace{1em}$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$

Equation $(1)$ will represent a circle if the coefficient of the $xy$ term is $0$.

i.e. $3 - p = 0$ $\implies$ $p = 3$

Substituting the value of $p$ in equation $(1)$ gives $\;\;$ $3x^2 + qy^2 - 6x - 24q = 0$

i.e. $x^2 + \dfrac{q}{3} y^2 - 2 x - 8q = 0$ $\;\;\; \cdots \; (3)$

Since equations $(2)$ and $(3)$ both represent the equation of circle, comparing the coefficients of the $y^2$ term gives

$\dfrac{q}{3} = 1$ $\implies$ $q = 3$

Substituting the value of $q$ in equation $(3)$ gives

$x^2 + y^2 - 2x -24 = 0$ $\;\;\; \cdots \; (4)$

Comparing equation $(4)$ with the standard equation of the circle [equation $(2)$] we get,

$2g = -2$ $\implies$ $g = - 1$, $\;\;\;$ $f = 0$, $\;\;\;$ $c = - 24$

Center of circle $= \left(-g, -f\right) = \left(1,0\right)$

Radius of circle $= \sqrt{g^2 + f^2 - c} = \sqrt{\left(-1\right)^2 + 0 + 24} = 5$ units