2-lines-through-the-point-A-5-1-are-tangent-to-the-circle-x-2-y-2-4x-6y-4-0-Find-the-equation-of-these-2-lines-

Question Number 31249 by Joel578 last updated on 04/Mar/18

2 lines through the point A (5, 1) are tangent

to the circle x^{2} + y^{2} - 4 x + 6 y + 4 = 0

Find the equation of these 2 lines

Commented by Joel578 last updated on 04/Mar/18

Equation of line through (5, 1)

y - 1 = m (x - 5)

y = m x - 5 m + 1

{(x - 2)}^{2} + {(y + 3)}^{2} = 9

Equation of tangent lines with slope m

y + 3 = m (x - 2) \pm 3 \sqrt{1 + m^{2}}

Substitute (5, 1)

4 = 3 (m \pm \sqrt{1 + m^{2}})

\frac{4}{3} - m = \pm \sqrt{1 + m^{2}}

\frac{16}{9} - \frac{8 m}{3} + m^{2} = 1 + m^{2}

\frac{16}{9} - 1 = \frac{8 m}{3}

16 - 9 = 24 m

m = \frac{7}{24}

Equation of tangent line

y = \frac{7}{24} x - \frac{11}{24}

Commented by Joel578 last updated on 04/Mar/18

I o n l y f o u n d 1 s o l u t i o n . {W h a t}^{'} s w r o n g

w i t h m y a n s w e r ?

Commented by ajfour last updated on 04/Mar/18

A l s o m = \infty

s o a n o t h e r t a n g e n t i s

x - 5 = 0 .

Commented by Joel578 last updated on 05/Mar/18

How do I know that another m = \infty

especially without using graphing tool ?

Commented by ajfour last updated on 05/Mar/18

w h e n y o u c a n c e l t e r m o f m^{2},

y o u s h o u l d r e a l i s e t h a t m = \pm \infty i s

o f c o u r s e t h e r o o t s . F o r e x a m p l e

m^{2} = {(m - 2)}^{2}

\Rightarrow m^{2} = m^{2} - 4 m + 4

t h i s e q . h a s r o o t s

m = \pm \infty a n d m = 1 .

Commented by Joel578 last updated on 05/Mar/18

S i r w o u l d y o u l i k e t o e x p l a i n i n d e t a i l e d w a y ?

I^{'} m s t i l l {d o n}^{'} t k n o w h o w c a n I s o l v e i f

I c h a n g e d y \Leftrightarrow x

Commented by Joel578 last updated on 05/Mar/18

O k a y . T h a n k y o u v e r y m u c h

Commented by MJS last updated on 05/Mar/18

{there}^{'} s no equation y = k x + d for

any vertical line

remember k = \tan α, α = 90 ° \Rightarrow \tan α = \pm \infty

in the given example, if you change

x \Leftrightarrow y {you}^{'} d get both solutions

Commented by MJS last updated on 05/Mar/18

{it}^{'} s enough to change the side of m

m (y - 1) = x - 5

x = m y - m + 5

x - 2 = m (y + 3) \pm 3 \sqrt{1 + m^{2}}

3 = 4 m \pm 3 \sqrt{1 + m^{2}}

3 - 4 m = \pm 3 \sqrt{1 + m^{2}}

9 - 24 m + 16 m^{2} = 9 + 9 m^{2}

7 m^{2} - 24 m = 0

m_{1} = 0

m_{2} = \frac{24}{7}

t_{1} : x = 0 y - 0 + 5 \Rightarrow x = 5

t_{2} : x = \frac{24}{7} y - \frac{24}{7} + 5 \Rightarrow x = \frac{24}{7} y + \frac{11}{7}

\Rightarrow y = \frac{7}{24} x - \frac{11}{24}

your method cannot show a

vertical tangent, mine cannot

show a horizontal one

so if one tangent is vertical

and the other {one}^{'} s horizontal,

we might need both

Commented by Joel578 last updated on 06/Mar/18

O k a y, t h a n k y o u f o r t h e e x p l a n a t i o n

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