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Question: If y = 2x + k touches $(x - 1)^2 + (y + 2)^2 = 1$, then find k and point of contact (given k > -2)....

If y = 2x + k touches (x−1)2+(y+2)2=1(x - 1)^2 + (y + 2)^2 = 1, then find k and point of contact (given k > -2).

A

k = 5−4\sqrt{5} - 4, Point of contact: (1−255,−2+55)\left(1 - \frac{2\sqrt{5}}{5}, -2 + \frac{\sqrt{5}}{5}\right)

B

k = −5−4-\sqrt{5} - 4, Point of contact: (1+255,−2−55)\left(1 + \frac{2\sqrt{5}}{5}, -2 - \frac{\sqrt{5}}{5}\right)

C

k = 5+4\sqrt{5} + 4, Point of contact: (1+255,−2+55)\left(1 + \frac{2\sqrt{5}}{5}, -2 + \frac{\sqrt{5}}{5}\right)

D

k = −5+4-\sqrt{5} + 4, Point of contact: (1−255,−2−55)\left(1 - \frac{2\sqrt{5}}{5}, -2 - \frac{\sqrt{5}}{5}\right)

Answer

k = 5−4\sqrt{5} - 4, Point of contact: (1−255,−2+55)\left(1 - \frac{2\sqrt{5}}{5}, -2 + \frac{\sqrt{5}}{5}\right)

Explanation

Solution

The circle (x−1)2+(y+2)2=1(x - 1)^2 + (y + 2)^2 = 1 has its center at C(1,−2)C(1, -2) and radius r=1r=1. The line y=2x+ky = 2x + k can be written as 2x−y+k=02x - y + k = 0. For the line to be tangent to the circle, the perpendicular distance from the center to the line must be equal to the radius. d=∣2(1)−(−2)+k∣22+(−1)2=∣4+k∣5d = \frac{|2(1) - (-2) + k|}{\sqrt{2^2 + (-1)^2}} = \frac{|4+k|}{\sqrt{5}} Setting d=r=1d=r=1: ∣4+k∣5=1  ⟹  ∣4+k∣=5\frac{|4+k|}{\sqrt{5}} = 1 \implies |4+k| = \sqrt{5} This gives two possibilities:

  1. 4+k=5  ⟹  k=5−44+k = \sqrt{5} \implies k = \sqrt{5}-4
  2. 4+k=−5  ⟹  k=−5−44+k = -\sqrt{5} \implies k = -\sqrt{5}-4

Given the condition k>−2k > -2: Since 5≈2.236\sqrt{5} \approx 2.236, k1=5−4≈2.236−4=−1.764k_1 = \sqrt{5}-4 \approx 2.236 - 4 = -1.764, which is greater than −2-2. k2=−5−4≈−2.236−4=−6.236k_2 = -\sqrt{5}-4 \approx -2.236 - 4 = -6.236, which is less than −2-2. Thus, we choose k=5−4k = \sqrt{5}-4.

The point of contact (x0,y0)(x_0, y_0) is the foot of the perpendicular from the center C(1,−2)C(1, -2) to the tangent line 2x−y+(5−4)=02x - y + (\sqrt{5}-4) = 0. Using the formula for the foot of the perpendicular: x0−x1A=y0−y1B=−Ax1+By1+CA2+B2\frac{x_0 - x_1}{A} = \frac{y_0 - y_1}{B} = -\frac{Ax_1 + By_1 + C}{A^2 + B^2} x0−12=y0−(−2)−1=−2(1)−1(−2)+(5−4)22+(−1)2\frac{x_0 - 1}{2} = \frac{y_0 - (-2)}{-1} = -\frac{2(1) - 1(-2) + (\sqrt{5}-4)}{2^2 + (-1)^2} x0−12=y0+2−1=−2+2+5−44+1\frac{x_0 - 1}{2} = \frac{y_0 + 2}{-1} = -\frac{2 + 2 + \sqrt{5}-4}{4+1} x0−12=y0+2−1=−55\frac{x_0 - 1}{2} = \frac{y_0 + 2}{-1} = -\frac{\sqrt{5}}{5} Equating the first part with the result: x0−12=−55  ⟹  x0−1=−255  ⟹  x0=1−255\frac{x_0 - 1}{2} = -\frac{\sqrt{5}}{5} \implies x_0 - 1 = -\frac{2\sqrt{5}}{5} \implies x_0 = 1 - \frac{2\sqrt{5}}{5} Equating the second part with the result: y0+2−1=−55  ⟹  y0+2=55  ⟹  y0=−2+55\frac{y_0 + 2}{-1} = -\frac{\sqrt{5}}{5} \implies y_0 + 2 = \frac{\sqrt{5}}{5} \implies y_0 = -2 + \frac{\sqrt{5}}{5} The point of contact is (1−255,−2+55)\left(1 - \frac{2\sqrt{5}}{5}, -2 + \frac{\sqrt{5}}{5}\right).