Find-the-equation-of-the-line-which-passes-through-the-point-3-5-and-is-tangent-to-the-circle-x-1-2-y-1-2-4-

Question Number 183044 by depressiveshrek last updated on 19/Dec/22

F i n d t h e e q u a t i o n o f t h e l i n e w h i c h

p a s s e s t h r o u g h t h e p o i n t (3, 5)

a n d i s t a n g e n t t o t h e c i r c l e

{(x - 1)}^{2} + {(y - 1)}^{2} = 4

Answered by mr W last updated on 19/Dec/22

s a y l i n e t h r o u g h (3, 5) i s :

x - 3 = k (y - 5)

\Rightarrow x - k y + 5 k - 3 = 0

\frac{∣ 1 - k + 5 k - 3 ∣}{\sqrt{1 + k^{2}}} = 2

\frac{∣ 2 k - 1 ∣}{\sqrt{1 + k^{2}}} = 1

(3 k - 4) k = 0

\Rightarrow k = 0 o r k = \frac{4}{3}

l i n e 1 : x - 3 = 0 \Rightarrow x = 3

l i n e 2 : x - 3 = \frac{4}{3} (y - 5) \Rightarrow y = \frac{3 x}{4} + \frac{11}{4}

Commented by mr W last updated on 19/Dec/22

Answered by MathAcer26 last updated on 19/Dec/22

The center is at (1, 1) with radius = 2

(3, 1) is a point in the circle .

Thus, x = 3 is a tangent to the circle passing

through (3, 5)

Answered by ajfour last updated on 20/Dec/22

l e t O r i g i n (1, 1)

c i r c l e x^{2} + y^{2} = 4

t a n g e n t h x + k y = 4

p a s s e s t h r o u g h (2, 4)

2 h + 4 k = 4 \Rightarrow h = 2 - 2 k

h^{2} + k^{2} = 4

\Rightarrow 4 {(1 - k)}^{2} + k^{2} = 4

5 k^{2} - 8 k = 0 \Rightarrow k = \frac{8}{5}, h = - \frac{6}{5}

e q . o f t a n g e n t i n O (0, 0)

h (x - 1) + k (y - 1) = 4

- 6 x + 8 y = 22

4 y = 3 x + 11

i s t h e t a n g e n t e q u a t i o n w i t h

O (0, 0) .

Answered by cortano1 last updated on 20/Dec/22

l e t t h e l i n e i s (a - 1) (x - 1) + (b - 1) (y - 1) = 4

a n d p a s s e s t h r o u g h t t h e p o i n t (3, 5)

\Rightarrow 2 (a - 1) + 4 (b - 1) = 4

\Rightarrow a - 1 + 2 b - 2 = 2; a = 5 - 2 b

(i i) {(a - 1)}^{2} + {(b - 1)}^{2} = 4

\Rightarrow {(4 - 2 b)}^{2} + {(b - 1)}^{2} = 4

\Rightarrow {\begin{cases} b = 1 \Rightarrow a = 3 \\ b = \frac{13}{5} \Rightarrow a = 5 - \frac{26}{5} = - \frac{1}{5} \end{cases}

s o t h e e q u a t i o n o f l i n e i s

\Rightarrow {\begin{cases} 2 (x - 1) = 4 \Rightarrow x = 3 \\ - \frac{6}{5} (x - 1) + \frac{8}{5} (y - 1) = 4 \end{cases}

\Rightarrow {\begin{cases} x = 3 \\ - 6 x + 6 + 8 y - 8 = 20 \end{cases}

\Rightarrow {\begin{cases} x = 3 \\ 3 x - 4 y + 11 = 0 \end{cases}

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