Question

Variable circle is described to pass through point (1,0) and tangent to the curve $y = \tan (\tan^{-1} x)$. The locus of the centre of the circle is a parabola whose :

• A. length of the latus rectum is $2\sqrt{2}$
• B. axis of symmetry has the equation $x + y = 1$
• C. vertex has the co-ordinates (3/4,1/4)
• D. none of these

Answer 1

Answer: (B), (C)

(b), (c)

Answer 2

The curve $y = \tan(\tan^{-1}x)$ simplifies to $y=x$. Let the center of the circle be $(h,k)$ and its radius be $r$. Since the circle passes through $(1,0)$, $r^2 = (h-1)^2 + k^2$. Since the circle is tangent to $y=x$ (or $x-y=0$), $r = \frac{|h-k|}{\sqrt{2}}$, so $r^2 = \frac{(h-k)^2}{2}$. Equating the expressions for $r^2$: $(h-1)^2 + k^2 = \frac{(h-k)^2}{2}$. This yields the locus equation $(h+k)^2 - 4h + 2 = 0$. Replacing $(h,k)$ with $(x,y)$, we get $(x+y)^2 - 4x + 2 = 0$. This is a parabola. Rotating the axes by $\theta = \pi/4$ using $X = \frac{x+y}{\sqrt{2}}$ and $Y = \frac{y-x}{\sqrt{2}}$, the locus transforms to $\left(X - \frac{1}{\sqrt{2}}\right)^2 = -\sqrt{2}\left(Y + \frac{\sqrt{2}}{4}\right)$. The axis of symmetry is $X = \frac{1}{\sqrt{2}}$, which translates to $x+y=1$. The vertex in the $(x,y)$ system is found to be $(3/4, 1/4)$. The length of the latus rectum is $|4a| = |-\sqrt{2}| = \sqrt{2}$.