Question

The equation of the circum-circle of the triangle formed by the lines$x=0,y=0,$ $\frac{x}{a}-\frac{y}{b}=1$, is

• A. $x^2+y^2+ax-by=0$
• B. $x^2+y^2-ax+by=0$
• C. $x^2+y^2-ax-by=0$
• D. $x^2+y^2+ax+by=0$

Answer 1

The correct option is B $x^2+y^2-ax+by=0$

Given line $bx-ay=ab$

Putting $y=0$ we get x-intercept as $x=a$

Putting $x=0$ we get y-intercept as $y=-b$

$△OAB$ is a right angled triangle as OA is perpendicular to OB

The circumcentre of a right angled triangle is always the midpoint of hypotenuse and radius will be half of the hypotenuse.

The coordinate of mid point of hypotenuse is $C(\frac{a}{2},-\frac{b}{2})$

The equation of a circle having $(\frac{a}{2},-\frac{b}{2})$ and radius be the $\sqrt{\frac{a^2+b^2}{4}}$

${(x-\frac{a}{2})}^2+{(y+\frac{b}{2})}^2=\frac{(a^2+b^2)}{4}$

$x^2-ax+\frac{a^2}{4}+y^2+by+\frac{b^2}{4}=\frac{a^2+b^2}{4}$

$x^2+y^2-ax+by=0$