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Question: Tangents are drawn from the point P(-2,0) to $y^2 = 8x$, radius of circle (s) that would touch these...

Tangents are drawn from the point P(-2,0) to y2=8xy^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(2−1)4(\sqrt{2}-1)

C

424\sqrt{2}

D

42+44\sqrt{2}+4

Answer

The possible radii are 4, 4(2−1)4(\sqrt{2}-1), and 4(2+1)4(\sqrt{2}+1). Options (A), (B), and (D) are correct.

Explanation

Solution

The parabola is y2=8xy^2 = 8x, so 4a=8  ⟹  a=24a=8 \implies a=2. The point is P(-2,0). The chord of contact is y(0)=2(2)(x+(−2))y(0) = 2(2)(x + (-2)), which simplifies to 0=4(x−2)0 = 4(x-2), so x=2x=2.

The tangents from P(-2,0) to y2=8xy^2 = 8x have equations y=mx+2my = mx + \frac{2}{m}. Substituting P(-2,0): 0=−2m+2m  ⟹  2m=2m  ⟹  m2=1  ⟹  m=±10 = -2m + \frac{2}{m} \implies 2m = \frac{2}{m} \implies m^2 = 1 \implies m = \pm 1. The tangents are y=x+2y = x+2 (or x−y+2=0x-y+2=0) and y=−x−2y = -x-2 (or x+y+2=0x+y+2=0).

Let the circle have center (h,k)(h,k) and radius rr. The circle touches x=2x=2, x−y+2=0x-y+2=0, and x+y+2=0x+y+2=0. The distance from (h,k)(h,k) to these lines must be rr.

  1. r=∣h−2∣r = |h-2|
  2. r=∣h−k+2∣12+(−1)2=∣h−k+2∣2r = \frac{|h-k+2|}{\sqrt{1^2+(-1)^2}} = \frac{|h-k+2|}{\sqrt{2}}
  3. r=∣h+k+2∣12+12=∣h+k+2∣2r = \frac{|h+k+2|}{\sqrt{1^2+1^2}} = \frac{|h+k+2|}{\sqrt{2}}

From (2) and (3): ∣h−k+2∣2=∣h+k+2∣2  ⟹  ∣h−k+2∣=∣h+k+2∣\frac{|h-k+2|}{\sqrt{2}} = \frac{|h+k+2|}{\sqrt{2}} \implies |h-k+2| = |h+k+2|. This implies h−k+2=h+k+2h-k+2 = h+k+2 (so k=0k=0) or h−k+2=−(h+k+2)h-k+2 = -(h+k+2) (so 2h=−4  ⟹  h=−22h = -4 \implies h=-2).

Case 1: k=0k=0. The center is (h,0)(h,0). r=∣h−2∣r = |h-2| and r=∣h+2∣2r = \frac{|h+2|}{\sqrt{2}}. ∣h−2∣=∣h+2∣2  ⟹  (h−2)2=(h+2)22|h-2| = \frac{|h+2|}{\sqrt{2}} \implies (h-2)^2 = \frac{(h+2)^2}{2} 2(h2−4h+4)=h2+4h+42(h^2-4h+4) = h^2+4h+4 2h2−8h+8=h2+4h+42h^2-8h+8 = h^2+4h+4 h2−12h+4=0h^2-12h+4 = 0 h=12±144−162=12±1282=12±822=6±42h = \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+42h = 6+4\sqrt{2}, r=∣(6+42)−2∣=∣4+42∣=4+42=4(2+1)r = |(6+4\sqrt{2})-2| = |4+4\sqrt{2}| = 4+4\sqrt{2} = 4(\sqrt{2}+1). If h=6−42h = 6-4\sqrt{2}, r=∣(6−42)−2∣=∣4−42∣=−(4−42)=42−4=4(2−1)r = |(6-4\sqrt{2})-2| = |4-4\sqrt{2}| = -(4-4\sqrt{2}) = 4\sqrt{2}-4 = 4(\sqrt{2}-1).

Case 2: h=−2h=-2. The center is (−2,k)(-2,k). r=∣−2−2∣=∣−4∣=4r = |-2-2| = |-4| = 4. Also, r=∣−2−k+2∣2=∣−k∣2=∣k∣2r = \frac{|-2-k+2|}{\sqrt{2}} = \frac{|-k|}{\sqrt{2}} = \frac{|k|}{\sqrt{2}}. So, 4=∣k∣2  ⟹  ∣k∣=424 = \frac{|k|}{\sqrt{2}} \implies |k| = 4\sqrt{2}. The radius is r=4r=4.

The possible radii are 4(2+1)4(\sqrt{2}+1), 4(2−1)4(\sqrt{2}-1), and 44.