let-the-cercle-x-1-2-y-3-2-9-and-the-point-A-4-1-vrrify-that-A-is-out-of-circle-and-determine-the-equation-of-two-tangentes-to-circle-wich-passes-by-point-A-

Question Number 77367 by msup trace by abdo last updated on 05/Jan/20

l e t t h e c e r c l e {(x + 1)}^{2} + {(y - 3)}^{2} = 9

a n d t h e p o i n t A (4, 1)

v r r i f y t h a t A i s o u t o f c i r c l e

a n d d e t e r m i n e t h e e q u a t i o n o f

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

p a s s e s b y p o i n t A .

Commented by mathmax by abdo last updated on 05/Jan/20

t h e p l a n i s p r o v i d e d w i t h o r t h o n o r m a l r e f e r e n c e (o, \vec{i}, \overset{\to)}{j}) .

Answered by jagoll last updated on 06/Jan/20

test point A (4, 1) \Rightarrow 5^{2} + {(- 2)}^{2} > 9

then A is out the circle

(2) let y = mx + n is tangent the circle

substitute point A \Rightarrow 1 = 4 m + n

n = 1 - 4 m . rewrite tangent line

y = mx + 1 - 4 m

distance of center circle to line equal

to radius \Rightarrow 3 =∣ \frac{- m + 1 - 4 m - 3}{\sqrt{1 + m^{2}}} ∣

3 \sqrt{1 + m^{2}} = ∣ - 5 m - 2 ∣

9 + {9 m}^{2} = {25 m}^{2} + 20 m + 4

{16 m}^{2} + 20 m - 5 = 0

m_{1} = \frac{1}{4} and m_{2} = - \frac{5}{4} . now you

get the tangent line .

Commented by msup trace by abdo last updated on 06/Jan/20

t h a n k s s i r .