Question

Let $A$ and $B$ be two fixed points then the locus of a point $C$ which moves so that $\left(\tan \angle BAC\right)\left(\tan \angle ABC\right)=1$ given that, $0<\angle BAC<\frac{\pi }{2},0<\angle ABC<\frac{\pi }{2}$ is

• A. Circle
• B. pair of straight line
• C. A point
• D. straight line

Answer 1

The correct option is A Circle

Given two fixed points A and B

Locus of C which i.e, $\left(\tan \angle BAC\right)\left(\tan \angle ABC\right)=1$
$\tan A\tan B=1\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$
$A+B=90°$
$\Rightarrow C=90°$

Let $C(x,y)$ be the point $C.$

$\therefore$ ABC is right angled at C

$A{B}^2=A{C}^2+B{C}^2$

$(x_2-x_1{)}^2+(y_2-y_1{)}^2=(x-x_1{)}^2+(y-y_1{)}^2+(x-x_2{)}^2+(y-y_2{)}^2$
$x_2^2-2x_2{x}_1+x_1^2+y_2^2-2y_2{y}_1+y_1^1=x^2-2_x{x}_1+x_1^2+y^2-2_y{y}_1+y_1^2+x^2-2_x{x}_2+x_2^2+y^2-2_y{y}_2+y$
$\Rightarrow 2x^2+2y^2-2(x_1+x_2-x_1{x}_2)x-2(y_1+y_2-y_1{y}_2)y=0$
$\therefore x^2+y^2-(x_1+x_2-x_1{x}_2)x-(y_1+y_2-y_1{y}_2)y=0$

$\therefore$ This is a circle