Question

The abscissa and ordinate of the points A and B are the roots of the equation $x^2+2ax+b=0$and $x^2+2cx+d=0$respectively, then equation of circle with AB as diameter is

• A. ${\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{+ 2ax + 2cy + b + d = 0}$
• B. ${\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{- 2ax - 2cy - b - d = 0}$
• C. ${\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{- 2ax - 2cy + b + d = 0}$
• D. ${\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{+ 2ax + 2cy - b - d = 0}$

Answer 1

The correct option is A ${\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{+ 2ax + 2cy + b + d = 0}$

$x^2+2ax+b=0letthepointsbeA(x_1,y_1)\&B(x_2,y_2)x=\frac{2a\pm \sqrt{(2a{)}^2-4\left(1\right)\left(b\right)}}{2}x=-a\pm \sqrt{a^2-b}{x}_1=-a+\sqrt{a^2-b}\&x_2=-a-\sqrt{a^2-b}$

$similarly,y_1=-c+\sqrt{c^2-d}{y}_2=-c-\sqrt{c^2-d}$
$Now,(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0(x-(-a+\sqrt{a^2-b}\left)\right)(x-(-a-\sqrt{a^2-b}\left)\right)+(y-(-c+\sqrt{c^2-d}\left)\right)(y-(-c-\sqrt{c^2-d}\left)\right)=0$
$aftersolving,x^2+y^2+2ax+2cy+b+d=0$