Question

A line is at a constant distance $c$ from the origin and meets the coordinates axes in $A$ and $B$. The locus of the centre of the circle passing through $O,A,B$ is

• A. $x^{-2}+y^{-2}=c^{-2}$
• B. $x^{-2}+y^{-2}=2c^{-2}$
• C. $x^{-2}+y^{-2}=3c^{-2}$
• D. $x^{-2}+y^{-2}=4c^{-2}$

Answer 1

Answer: (D)

$x^{-2}+y^{-2}=4c^{-2}$

Let the equation of line be
$\frac{x}{a}+\frac{y}{b}=1$
where $a$ and $b$ are the x-intercept and y-intercept.
Then the coordinates of A and B are $(a,0)$ and $(0,b)$
Distance of origin to the line is $c$
$\Rightarrow c=\frac{|-1|}{\sqrt{(\frac{1}{a}{)}^2+(\frac{1}{b}{)}^2}}$
$\Rightarrow \frac{1}{c}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}$
$\Rightarrow \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}$ ....(1)
Let the center of circle through O, A, B be $(h,k)$
Points $O(0,0),A(a,0),(0,b)$ forms a right triangle.
So, the center of circle is the mid-point of AB i.e.$(\frac{a}{2},\frac{b}{2})$
$\Rightarrow h=\frac{a}{2},k=\frac{b}{2}$
$\Rightarrow a=2h,b=2k$
Substitute this value in (1), we get
$\frac{1}{c^2}=\frac{1}{4h^2}+\frac{1}{4k^2}$
$\Rightarrow \frac{4}{c^2}=\frac{1}{h^2}+\frac{1}{k^2}$
$\Rightarrow 4c^{-2}=x^{-2}+y^{-2}$ (Replacing $h,k$ by $x,y$ )