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+1
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x+ky=1 is a tangent to the circle (x+3)^2 + (y-2)^2 = 16, find the value of K.

Jun 5, 2019

#3
+25228
+4

x+ky=1 is a tangent to the circle (x+3)^2 + (y-2)^2 = 16, find the value of K.

$$\begin{array}{|rcll|} \hline \mathbf{(x+3)^2 + (y-2)^2} &=& \mathbf{16} \quad |\quad x+ky=1 \text{ or } \mathbf{x = 1-ky} \\ ( 1-ky+3)^2 + (y-2)^2 &=& 16 \\ ( 4-ky)^2 + (y-2)^2 &=& 4^2 \\ 4^2-8ky+k^2y^2+y^2-4y+4 &=& 4^2 \\ -8ky+k^2y^2+y^2-4y+4 &=& 0 \\ y^2(1+k^2)-4y(1+2k)+4 &=& 0 \\\\ y &=& \dfrac{4(1+2k)\pm \sqrt{4^2(1+2k)^2-4(1+k^2)4} }{2(1+k^2)} \\\\ y &=& \dfrac{4(1+2k)\pm 4 \sqrt{(1+2k)^2-(1+k^2)} }{2(1+k^2)} \\\\ y &=& \dfrac{4(1+2k)\pm 4 \overbrace{\sqrt{\color{red}(1+2k)^2-(1+k^2)}}^{=0~(\text{tangent}) } }{2(1+k^2)} \\\\ \mathbf{\color{red}(1+2k)^2-(1+k^2)} &=& \mathbf{0}\\ 1+4k+4k^2-1-k^2 &=& 0 \\ 4k+4k^2 -k^2 &=& 0 \\ 3k^2 +4k &=& 0 \\ k(3k +4 ) &=& 0 \\ && 1.)~ \mathbf{k = 0} \\ && 2.)~ 3k +4 = 0 \\ && \quad ~ \mathbf{k = -\dfrac{4}{3}} \\ \hline \end{array}$$

The value of k is $$\mathbf{0}$$ or $$\mathbf{-\dfrac{4}{3}}$$

Jun 5, 2019

#1
+1

$$P(1;2)\\ k\ →\ \frac{1}{ ∞ }$$

.
Jun 5, 2019
#1
0

$$P(1;2)\\ k\ →\ \frac{1}{ ∞ }$$

Guest Jun 5, 2019
#3
+25228
+4

x+ky=1 is a tangent to the circle (x+3)^2 + (y-2)^2 = 16, find the value of K.

$$\begin{array}{|rcll|} \hline \mathbf{(x+3)^2 + (y-2)^2} &=& \mathbf{16} \quad |\quad x+ky=1 \text{ or } \mathbf{x = 1-ky} \\ ( 1-ky+3)^2 + (y-2)^2 &=& 16 \\ ( 4-ky)^2 + (y-2)^2 &=& 4^2 \\ 4^2-8ky+k^2y^2+y^2-4y+4 &=& 4^2 \\ -8ky+k^2y^2+y^2-4y+4 &=& 0 \\ y^2(1+k^2)-4y(1+2k)+4 &=& 0 \\\\ y &=& \dfrac{4(1+2k)\pm \sqrt{4^2(1+2k)^2-4(1+k^2)4} }{2(1+k^2)} \\\\ y &=& \dfrac{4(1+2k)\pm 4 \sqrt{(1+2k)^2-(1+k^2)} }{2(1+k^2)} \\\\ y &=& \dfrac{4(1+2k)\pm 4 \overbrace{\sqrt{\color{red}(1+2k)^2-(1+k^2)}}^{=0~(\text{tangent}) } }{2(1+k^2)} \\\\ \mathbf{\color{red}(1+2k)^2-(1+k^2)} &=& \mathbf{0}\\ 1+4k+4k^2-1-k^2 &=& 0 \\ 4k+4k^2 -k^2 &=& 0 \\ 3k^2 +4k &=& 0 \\ k(3k +4 ) &=& 0 \\ && 1.)~ \mathbf{k = 0} \\ && 2.)~ 3k +4 = 0 \\ && \quad ~ \mathbf{k = -\dfrac{4}{3}} \\ \hline \end{array}$$

The value of k is $$\mathbf{0}$$ or $$\mathbf{-\dfrac{4}{3}}$$

heureka Jun 5, 2019