<p>Find the locus of the feet of the perpendiculars from the point \((h, k)\) to the tangents to the circle \(x^2 + y^2 = 2ax\).</p>
Step-by-Step Solution
Key Concept: The foot of perpendicular from point (h,k) to a tangent of the circle lies on a circle whose diameter is the line joining (h,k) to points on the original circle. Use the property that the tangent at any point on the circle is perpendicular to the radius at that point.
<p><strong>Step 1:</strong> Rewrite the circle in standard form: $(x-a)^2 + y^2 = a^2$, with center C(a, 0) and radius a.</p><p><strong>Step 2:</strong> For any point P(x,y) on the locus, it is the foot of perpendicular from A(h,k) to a tangent. If the tangent touches the circle at point T, then CT ⊥ tangent, and AP ⊥ tangent, so CT || AP.</p><p><strong>Step 3:</strong> Since P lies on the tangent at T and AP ⊥ tangent, we have CP ⊥ AP. Therefore, angle CPA = 90°.</p><p><strong>Step 4:</strong> The locus is a circle with CA as diameter. Using the condition that CP ⊥ AP for any foot of perpendicular P:</p><p>$(x-a)(h-x) + (y-0)(k-y) = 0$</p><p>$∴ x^2 + y^2 - (a+h)x - ky + ah = 0$</p><p>This is the required locus—a circle passing through point C(a,0) with diameter endpoints at A(h,k) and C(a,0).</p>
Correct Answer: ((ax-y^2)(k-y) + y(x-a)(k-y))^2 + ... (locus equation)