Question:

Step 1: Define the point and the circle, and state the general formula for the combined equation of tangents. Let the external point from which the tangents are drawn be $P(h,k)$. The equation of the circle is $x^2 + y^2 = 1$. The combined equation of the pair of tangents from an external point $P(h,k)$ to a circle $S = x^2+y^2-r^2=0$ is given by $T^2 = SS_1$, where $S_1 = h^2+k^2-r^2$ and $T = hx+ky-r^2$. For the given circle $x^2+y^2-1=0$, we have $r^2=1$. Thus, $S_1 = h^2+k^2-1$ and $T = hx+ky-1$. Step 2: Write down the specific combined equation of tangents for this problem. The combined equation of tangents from $P(h,k)$ to the circle $x^2+y^2=1$ is: $$ (hx+ky-1)^2 = (x^2+y^2-1)(h^2+k^2-1) $$ Step 3: Substitute $x=1$ to find the points of intersection with the line $x=1$, and form a quadratic equation in $y$. The tangents make an intercept on the line $x=1$. Substitute $x=1$ into the combined equation: $$ (h(1)+ky-1)^2 = ((1)^2+y^2-1)(h^2+k^2-1) $$ $$ (h-1+ky)^2 = (1+y^2-1)(h^2+k^2-1) $$ $$ ((h-1)+ky)^2 = y^2(h^2+k^2-1) $$ Expand the left side: $$ (h-1)^2 + 2k(h-1)y + k^2y^2 = y^2(h^2+k^2-1) $$ Rearrange the terms to form a quadratic equation in $y$: $$ y^2(k^2 - (h^2+k^2-1)) + 2k(h-1)y + (h-1)^2 = 0 $$ $$ y^2(k^2 - h^2 - k^2 + 1) + 2k(h-1)y + (h-1)^2 = 0 $$ $$ y^2(1-h^2) + 2k(h-1)y + (h-1)^2 = 0 $$ Factor out $(1-h)$ from the terms (assuming $h \ne 1$): $$ y^2(1-h)(1+h) + 2k(h-1)y + (1-h)^2 = 0 $$ Divide by $(1-h)$: $$ y^2(1+h) - 2ky - (h-1) = 0 $$ This is the quadratic equation whose roots $y_1, y_2$ are the $y$-coordinates of the intersection points of the tangents with the line $x=1$. Step 4: Use the given condition that the intercept on $x=1$ is 2 units. The intercept of 2 units on the line $x=1$ means that the difference between the $y$-coordinates of the intersection points is 2. So, $|y_1 - y_2| = 2$. For a quadratic equation $Ay^2+By+C=0$, the difference between the roots is given by $|y_1 - y_2| = \frac{\sqrt{B^2-4AC}}{|A|}$. From our quadratic $y^2(h+1) - 2ky - (h-1) = 0$, we have $A = (h+1)$, $B = -2k$, and $C = -(h-1)$. So, we have: $$ 2 = \frac{\sqrt{(-2k)^2 - 4(h+1)(-(h-1))}}{|h+1|} $$ $$ 2 = \frac{\sqrt{4k^2 + 4(h+1)(h-1)}}{|h+1|} $$ $$ 2 = \frac{\sqrt{4k^2 + 4(h^2-1)}}{|h+1|} $$ Step 5: Solve for the relationship between $h$ and $k$. Square both sides of the equation: $$ 4 = \frac{4k^2 + 4(h^2-1)}{(h+1)^2} $$ Divide by 4: $$ 1 = \frac{k^2 + (h^2-1)}{(h+1)^2} $$ Multiply by $(h+1)^2$: $$ (h+1)^2 = k^2 + h^2 - 1 $$ $$ h^2 + 2h + 1 = k^2 + h^2 - 1 $$ Subtract $h^2$ from both sides: $$ 2h + 1 = k^2 - 1 $$ $$ k^2 = 2h + 2 $$ $$ k^2 = 2(h+1) $$ Step 6: Express the locus and identify its type. Replacing $(h,k)$ with $(x,y)$ to find the locus of $P$, we get: $$ y^2 = 2(x+1) $$ This is the equation of a parabola. The final answer is $\boxed{\text{parabola}}$.