Question

Values of $p$ and $q$ for which equation $(x+y-41{)}^2-(x+7y-7\left)\right(px+qy+1)=0$ represents a circle are

• A. $p=-7/25,q=-1/25$
• B. $p=7/25,q=1/25$
• C. $p=1/25,q=7/25$
• D. $p=-1/25,q=-7/25$

Answer 1

The correct option is B $p=7/25,q=1/25$

On expanding, the equation becomes
$(1-p)x^2+(1-7q)y^2+(2-q-7p)xy+(-82-1+7p)x+(-82-7+7p)y+1688=0$
Now, we equate the coefficients of $x^2$ and $y^2$ also of x and y, and make the coefficient of $xy=0$ since the equation represents a circle.
Hence,
$1-p=1-7q$
$2-q-7p=0$
On solving we get,
$p=\frac{7}{25}$, $q=\frac{1}{25}$