Question

A chord of the circle $x^2+y^2-4x-6y=0$ passing through the origin subtends an angle $ta{n}^{-1}\left(\frac{7}{4}\right)$ at the point where the circle meets positive y-axis. Equation of the chord is

• A. 2x + 3y = 0
• B. x + 2y = 0
• C. x – 2y = 0
• D. 2x – 3y = 0

Answer 1

The correct option is C x – 2y = 0


The given circle passes through the origin O and meets the positive Y-axis at B(0, 6). Let OP be the chord of the circle passing through the origin subtending an angle $\theta$ at B, where $tan\theta =\frac{7}{4}$
$\Rightarrow \angle OBP=\theta$
Equation of the tangent OT at O to the given circle is 2x + 3y = 0
$\Rightarrow slopeofthetangent=-\frac{2}{3}$
So that, if $\angle XOT=\alpha$, $tan\alpha =\frac{2}{3}$
From geometry, $\angle POT=\angle OBP=\theta \Rightarrow \angle POT=\theta -\alpha$
and $tan(\theta -\alpha )=\frac{tan\theta -tan\alpha }{1+tan\theta tan\alpha }=\frac{\frac{7}{4}-\frac{2}{3}}{1+\frac{7}{4}\times \frac{2}{3}}=\frac{13}{26}=\frac{1}{2}$
Hence the equation of OP is $y=xtan(\theta -\alpha )\Rightarrow x-2y=0$