Question

If the equation $p{x}^2+(2-q)xy+3y^2-6qx+30y+6q=0$ represents a circle, then which of the following is/are correct?

• A. $3p+2q=0$
• B. $2p-3q=0$
• C. $(p-1)(q-1)=2$
• D. $3p-2q=0$

Answer 1

Answer: (B), (C)

The correct options are
B $2p-3q=0$
C $(p-1)(q-1)=2$

The general equation of second degree is :
$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$
Given : $p{x}^2+(2-q)xy+3y^2-6qx+30y+6q=0$
Comparing both,
$a=p,b=3,h=\frac{2-q}{2},f=15,g=3q,c=6q$
For circle, coefficient of $xy=0$
and coefficient of $x^2=$ coefficient of $y^2$
$\Rightarrow 2-q=0\Rightarrow q=2$
and $p=3$

Now, $\Delta =abc+2fgh-a{f}^2-b{g}^2-c{h}^2$
$=18pq+2\left(15\right)\left(3q\right)\left(\frac{2-q}{2}\right)-p(15{)}^2-3(3q{)}^2-6q{\left(\frac{2-q}{2}\right)}^2$
$=-675$
$\Rightarrow \Delta \ne 0$