JEE Main 2022 (Online) 26th July Evening ShiftQuestion 1
Let the abscissae of the two points $P$ and $Q$ on a circle be the roots of $x^2-4x-6=0$ and the ordinates of $P$ and $Q$ be the roots of $y^2+2y-7=0$. If $PQ$ is a diameter of the circle $x^2+y^2+2ax+2by+c=0$, then the value of $(a+b-c)$ is
Circle
Solution
Concept. Use the diameter form of the circle and Vieta's formulas for the coordinates. Formulas used. (1) Circle on diameter $PQ$: $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$(2) Vieta: $x_1+x_2=4$, $x_1x_2=-6$; $y_1+y_2=-2$, $y_1y_2=-7$Step 1 - Expand (1). $$x^2+y^2-(x_1+x_2)x-(y_1+y_2)y+x_1x_2+y_1y_2=0.$$ Step 2 - Use (2). $x^2+y^2-4x+2y-13=0$. Step 3 - Compare. $2a=-4\Rightarrow a=-2$; $2b=2\Rightarrow b=1$; $c=-13$. Step 4 - Value. $a+b-c=-2+1+13=12$. Answer: A