Question

If OA and OB are equal perpendicular chords of the circles $x^2+y^2-2x+4y=0$, then equation of OA and OB are where O is origin.

• A. $3x+y=0$ and $3x-y=0$
• B. $3x+y=0$ or $3y-x=0$
• C. $x+3y=0$ and $y-3x=0$
• D. $x+y=0$ or $x-y=0$

Answer 1

Answer: (C)

$x+3y=0$ and $y-3x=0$

Let the equation of the chord OA of a circle
$x^2+y^2-2x+4y=0$ ......... (i)
be $y=mx$ ........ (ii)

Solving (i) and (ii), we get
$\Rightarrow x^2+m^2{x}^2-2x+4mx=0$

$\Rightarrow (1+m^2)x^2-(2-4m)x=0$

$\Rightarrow x=0$ and $x=\frac{2-4m}{1+m^2}$

Hence, the points of intersection are
$(0,0)$ and $A(\frac{2-4m}{1+m^2},\frac{m(2-4m)}{1+m^2})$

$\Rightarrow O{A}^2={\left(\frac{2-4m}{1+m^2}\right)}^2(1+m^2)$

$=\frac{(2-4m{)}^2}{1+m^2}$
Since $△OAB$ is an isosceles right-angled triangle $O{A}^2=\frac{1}{2}A{B}^2$

$AB$ is a diameter of the given circle

$O{A}^2=10$

$\Rightarrow \frac{(2-4m{)}^2}{1+m^2}=10$

$\Rightarrow 4-16+16m^2=10(1+m{)}^2$

$\Rightarrow 3m^2-8m-3=0$

$\Rightarrow m=3$ or $-\frac{1}{3}$

Hence, the required equations are $y=3x$ or $x+3y=0$