Question

Tangents are drawn from the point P(-2,0) to $y^2 = 8x$, radius of circle (s) that would touch these tangents and the corresponding chord of contact can be equal to

• A. 4
• B. 4($\sqrt{2}$-1)
• C. 4$\sqrt{2}$
• D. 4$\sqrt{2}$+4

Answer 1

Answer: (B), (D)

(B), (D)

Answer 2

The given parabola is $y^2 = 8x$, which implies $4a=8$, so $a=2$. The point P is $(-2, 0)$.

The equation of the tangents to $y^2=4ax$ with slope $m$ is $y = mx + \frac{a}{m}$. Since the tangents pass through P$(-2,0)$: $0 = m(-2) + \frac{2}{m} \implies -2m + \frac{2}{m} = 0 \implies 2m = \frac{2}{m} \implies m^2 = 1 \implies m = \pm 1$. The equations of the tangents are: For $m=1$: $y = x + 2 \implies x - y + 2 = 0$. For $m=-1$: $y = -x - 2 \implies x + y + 2 = 0$.

The equation of the chord of contact from $(x_1, y_1)$ to $y^2=4ax$ is $yy_1 = 2a(x+x_1)$. For P$(-2, 0)$: $y(0) = 2(2)(x + (-2)) \implies 0 = 4(x-2) \implies x=2$. The chord of contact is the line $x=2$.

Let the circle have center $(h, k)$ and radius $r$. The distance from $(h, k)$ to the tangent $x-y+2=0$ is $\frac{|h-k+2|}{\sqrt{2}}$. The distance from $(h, k)$ to the tangent $x+y+2=0$ is $\frac{|h+k+2|}{\sqrt{2}}$. These distances must be equal to $r$. $\frac{|h-k+2|}{\sqrt{2}} = \frac{|h+k+2|}{\sqrt{2}} \implies |h-k+2| = |h+k+2|$. This implies either $h-k+2 = h+k+2$ (which gives $k=0$) or $h-k+2 = -(h+k+2)$ (which gives $2h=-4$, so $h=-2$).

The distance from $(h, k)$ to the chord of contact $x=2$ is $|h-2|$. This must also be equal to $r$. So, $r = |h-2|$.

Case 1: $k=0$. The center is $(h, 0)$. $r = |h-2|$. The distance to the tangents is $\frac{|h-0+2|}{\sqrt{2}} = \frac{|h+2|}{\sqrt{2}}$. So, $r = \frac{|h+2|}{\sqrt{2}}$. Equating the expressions for $r$: $|h-2| = \frac{|h+2|}{\sqrt{2}}$. Squaring both sides: $(h-2)^2 = \frac{(h+2)^2}{2} \implies 2(h^2-4h+4) = h^2+4h+4 \implies 2h^2-8h+8 = h^2+4h+4 \implies h^2-12h+4 = 0$. Using the quadratic formula, $h = \frac{12 \pm \sqrt{144 - 16}}{2} = \frac{12 \pm \sqrt{128}}{2} = \frac{12 \pm 8\sqrt{2}}{2} = 6 \pm 4\sqrt{2}$.

If $h = 6+4\sqrt{2}$, then $r = |(6+4\sqrt{2})-2| = |4+4\sqrt{2}| = 4+4\sqrt{2}$. This matches option (D). If $h = 6-4\sqrt{2}$, then $r = |(6-4\sqrt{2})-2| = |4-4\sqrt{2}| = 4\sqrt{2}-4$. This matches option (B).

Case 2: $h=-2$. The center is $(-2, k)$. $r = |h-2| = |-2-2| = |-4| = 4$. This matches option (A). The distance to the tangents is $\frac{|-2-k+2|}{\sqrt{2}} = \frac{|-k|}{\sqrt{2}} = \frac{|k|}{\sqrt{2}}$. So, $r = \frac{|k|}{\sqrt{2}} \implies 4 = \frac{|k|}{\sqrt{2}} \implies |k| = 4\sqrt{2}$. The points of contact on the parabola are $(2, 4)$ and $(2, -4)$. The chord of contact is the segment on the line $x=2$ from $y=-4$ to $y=4$. For the circle to touch the chord of contact segment, the point of tangency on $x=2$ must lie on this segment. The point of tangency is $(2, k)$. Thus, we must have $-4 \le k \le 4$. However, we found $|k| = 4\sqrt{2} \approx 5.657$, which is outside the range $[-4, 4]$. Therefore, a circle with radius $r=4$ cannot touch the chord of contact segment. If the question meant the line $x=2$, then $r=4$ would be valid. Given the context of "chord of contact" of tangents, it usually refers to the line segment connecting the points of tangency.

Option (C) $4\sqrt{2}$: If $r=4\sqrt{2}$, then $|h-2|=4\sqrt{2} \implies h = 2 \pm 4\sqrt{2}$. Also, the distance to the tangents must be $4\sqrt{2}$. If $h=-2$, $r=4$, which is not $4\sqrt{2}$. If $k=0$, $r=\frac{|h+2|}{\sqrt{2}} \implies 4\sqrt{2} = \frac{|h+2|}{\sqrt{2}} \implies |h+2|=8 \implies h+2=\pm 8 \implies h=6$ or $h=-10$. Neither of these match $h=2 \pm 4\sqrt{2}$.

The valid radii that touch the tangents and the chord of contact segment are $4\sqrt{2}-4$ and $4+4\sqrt{2}$.