Circles
Tangents and Chords
Grade 11
Question:
<p>The locus of points of intersection of the tangents to $x^2 + y^2 = a^2$ at the extremities of a chord of circle $x^2 + y^2 = a^2$ which touches the circle $x^2 + y^2 - 2ax = 0$ is/are:</p>
<p>(a) $y^2 = a(a - 2x)$</p>
<p>(b) $x^2 = a(a - 2y)$</p>
<p>(c) $x^2 + y^2 = (x - a)^2$</p>
<p>(d) $x^2 + y^2 = (y - a)^2$</p>
Step-by-Step Solution
Key Concept: The locus is found by requiring the chord (whose tangent endpoints are being considered) to be tangent to the given inner circle.
<p><strong>Analysis:</strong> The circle $x^2 + y^2 - 2ax = 0$ has centre $(a, 0)$ and radius $a$. It touches the circle $x^2 + y^2 = a^2$. A chord of $x^2 + y^2 = a^2$ that touches $x^2 + y^2 - 2ax = 0$ can be parameterized. If the chord has equation $lx + my = a^2/p$ (for some parameter $p$), the tangents at its extremities meet at the pole of the chord with respect to the circle, which lies on the polar. The condition that the chord touches the inner circle $x^2 + y^2 - 2ax = 0$ gives the locus. Working through the geometry: if $(h,k)$ is the intersection of tangents to the chord, the chord equation is $hx + ky = a^2$. This chord touches $x^2 + y^2 - 2ax = 0$ when the distance from $(a,0)$ to $hx + ky = a^2$ equals $a$: $$\frac{|ah - a^2|}{\sqrt{h^2+k^2}} = a$$. This simplifies to $y^2 = a(a-2x)$ where $(x,y) = (h,k)$.</p><p>∴ Answer is (a).</p>
Correct Answer: a