Question

Equation of the straight line which meets the circle $x^2+y^2=a^2$ at points which are at a distance d from a point $A(\alpha ,\beta )$ on the circle is

• A. $2\alpha x+2\beta y=2a^2-d^2$
• B. $2\alpha x-2\beta y=2a^2+d^2$
• C. $2\alpha x+2\beta y=2a^2+d^2$
• D. $2\alpha x+2\beta y+2a^2=d^2$

Answer 1

Answer: (A), (D)

$2\alpha x+2\beta y=2a^2-d^2$
$2\alpha x+2\beta y+2a^2=d^2$

Let $M$ and $N$ be the points on the circle $x^2+y^2=a^2$ such that $AM=AN=d$ (from figure)

The line joining $A(\alpha ,\beta )$ and $O(0,0),$(the center of the circle) meets the chord $MN$ at $B$ the mid-point of $MN$.

Now, the slope of $OA$ is $\frac{\beta }{\alpha }$

So that the slope of $MN$ is $\frac{\alpha }{\beta }$

An equation of $MN$ is $\alpha x+\beta y=k.$

Then, since $\frac{\alpha }{\beta }$ lies on the given circle, we have

$OB=\left|\frac{-k}{\sqrt{{\alpha }^2+{\beta }^2}}\right|=\left|\frac{-k}{a}\right|$

$AB=a\pm \frac{k}{a},B{M}^2=O{M}^2-O{B}^2=a^2\frac{k^2}{a^2}$

and $A{M}^2=A{B}^2+B{M}^2$

$\Rightarrow d^2={(a\pm \frac{k}{a})}^2+a^2-\frac{k^2}{a^2}\Rightarrow d^2=2a^2\pm 2k$

$k=\pm \frac{2a^2-d^2}{2}$