Question

If the abscissas and ordinates of two points $P$ and $Q$ are the roots of the equations $x^2-7x+10=0$ and $x^2+7x+12=0$ respectively, then the equation of the circle with $PQ$ as diameter is

• A. $x^2+y^2+5x+2y+10=0$
• B. $x^2+y^2-4x-3y+12=0$
• C. $x^2+y^2-7x+7y+22=0$
• D. $x^2+y^2-5x+2y+10=0$

Answer 1

The correct option is C $x^2+y^2-7x+7y+22=0$

Let the points be $P=(x_1,y_1),Q=(x_2,y_2)$
$x^2-7x+10=0$ has $x_1,x_2$ as roots, so
$(x-5)(x-2)=0\Rightarrow x=5,2$
$x^2+7x+12=0$ has $y_1,y_2$ as roots, so
$(x+4)(x+3)=0\Rightarrow x=-4,-3$

Therefore, equation of circle in diameter form is
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0\Rightarrow (x-5)(x-2)+(y+3)(y+4)=0\therefore x^2+y^2-7x+7y+22=0$


Alternate solution :
Given equations are
$x^2-7x+10=0y^2+7y+12=0$
Adding both, we get the required equation of circle,
$x^2+y^2-7x+7y+22=0$