Question

The locus of the point of intersection of the tangents at the extremeties of a chord of the circle $x^2+y^2 = a^2$ which touches the circle $x^2 + y^2 - 2ax = 0$ passes through the point

• A. $(a/2,0)$
• B. $(0,a/2)$
• C. $(a, 0)$
• D. $(0, 0)$

Answer 1

Answer: (A)

$(a/2,0)$

Answer 2

Equation can be rewritten as
$( x - a)^2 + y^2 = a^2$
Any point on the circle is $(a + a \,cos\, \theta, a \,sin\, \theta)$
$\therefore$ Equation of tangent at $(a + a \,cos\,\theta, a \,sin\,\theta)$ is
$x(a + a \,cos\,\theta) + y(a \,sin\,\theta)$
$-a(x + a + a \,cos\,\theta) = 0$
$\Rightarrow ax\,cos\,\theta + ay \,sin\,\theta - a^2(1 + cos\,\theta)\,\, ,..(i)$
Equation of first circle is
$x^2 + y^2 = a^2 ...(ii)$
Let E $(i)$ meets the first circle at $P$ and $Q$ and the tangents at $P$ and $Q$ to the second circle intersected at $(h, k)$, then E $(i)$ is the chord of contact of $(h, k)$ with respect to the circle $(ii)$ and thus equation is
$h x + k y - a^2 = 0 \,\,...(iii)$
Eqs. $(i)$ and $(iii)$ represents the same line.
$\therefore \frac{h}{a\,cos\,\theta} = \frac{k}{a\,sin\,\theta} = \frac{a^2}{a^2(1 + cos\,\theta)}$
$\Rightarrow \frac{h}{a} = \frac{cos\,\theta}{1 + cos\,\theta}, \frac{k}{a} = \frac{sin\,\theta}{1 + cos\,\theta}$
\Rightarrow (\frac{h}{a})^2 + (\frac{k}{a})^2 = \frac{cos^2\,\theta + sin^\,\theta}{(1 + cos\,\theta)^2}
$= \frac{1}{4[cos^2 \frac{\theta}{2}]^2}$
Then, $\left(\frac{a/2}{a}\right)^{2} + 0^{2} = \frac{1}{4} \left( 1 +0\right)^{2}$ $\Rightarrow \frac{1}{4} = \frac{1}{4}$
Hence, required point is $\left( \frac{a}{2} , 0\right)$