Find-the-equation-of-a-circle-which-touches-x-axis-and-the-line-y-x-in-the-1-st-quadrant-Determine-its-centre-and-radius-if-it-touches-the-line-y-x-4-

Question Number 116093 by Engr_Jidda last updated on 30/Sep/20

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

x - a x i s a n d t h e l i n e y = x i n t h e 1^{s t} q u a d r a n t .

D e t e r m i n e i t s c e n t r e a n d r a d i u s i f i t t o u c h e s t h e l i n e y + x = 4 .

Answered by john santu last updated on 01/Oct/20

l e t P (2, b) b e a c e n t r e p o i n t a c i r c l e

t o u c h e s x - a x i s; l i n e y = x a n d y + x = 4

(i) b = \frac{∣ 2 - b ∣}{\sqrt{2}} \Rightarrow 2 b^{2} = 4 - 4 b + b^{2}

b^{2} + 4 b - 4 = 0; {(b + 2)}^{2} = 8

b = 2 \sqrt{2} - 2

(i i) b = \frac{∣ 2 + b - 4 ∣}{\sqrt{2}} \Rightarrow b = 2 \sqrt{2} - 2

s o w e h a v e \to {\begin{cases} P (2, 2 \sqrt{2} - 2) \\ r = 2 \sqrt{2} - 2 \end{cases}

e q u a t i o n o f a c i r c l e i s

{(x - 2)}^{2} + {(y - 2 \sqrt{2} + 2)}^{2} = {(2 \sqrt{2} - 2)}^{2}

{(x - 2)}^{2} + {(y - 2 \sqrt{2} + 2)}^{2} = 12 - 8 \sqrt{2}

Commented by bemath last updated on 01/Oct/20

Commented by bemath last updated on 01/Oct/20

great mr^{″} a farmer {math}^{″}

Answered by 1549442205PVT last updated on 02/Oct/20

a) Suppose the equation of the circle C is

{(x - a)}^{2} + {(y - b)}^{2} = R^{2} . Since C touches

the lines y = 0 and y = x, we infer the

equations {(x - a)}^{2} + b^{2} = R^{2} (1)

and {(x - a)}^{2} + {(x - b)}^{2} = R^{2} (2) has unique

root

(1) \Leftrightarrow x^{2} - 2 ax + a^{2} + b^{2} - R^{2} = 0

with Δ^{'} = a^{2} - a^{2} - b^{2} + R^{2} = R^{2} - b^{2}

(2) \Leftrightarrow {2 x}^{2} - 2 (a + b) x + a^{2} + b^{2} - R^{2} = 0

with Δ^{'} = {(a + b)}^{2} - {2 a}^{2} - {2 b}^{2} + {2 R}^{2}

= {2 R}^{2} - {(a - b)}^{2} . Therefore, (1) (2) has

unique root if and only if

{\begin{cases} R^{2} - b^{2} = 0 \\ {2 R}^{2} - {(a - b)}^{2} = 0 \end{cases} \Leftrightarrow {\begin{cases} b = \pm R \\ a = \pm R \pm R \sqrt{2} \end{cases} (*)

Hence, the equations of C are : (four cases)

i) b = R \Rightarrow a = R (1 \pm \sqrt{2}) gives two the circles family :

{(x - R (1 + \sqrt{2}))}^{2} + {(y - R)}^{2} = R^{2} .

{(x - R (1 - \sqrt{2}))}^{2} + {(y - R)}^{2} = R^{2}

ii) b = - R \Rightarrow a = R (- 1 \pm \sqrt{2}) gives two the circles family :

{[x - R (- 1 + \sqrt{2})]}^{2} + {(y + R)}^{2} = R^{2}

[{(x - R (- 1 - \sqrt{2})]}^{2} + {(y + R)}^{2} = R^{2}

Example, R = 1, b = 1, a = 1 + \sqrt{2} \dots .

b) Now C touches to the x + y = 4 if and

only if the equation {(x - a)}^{2} + {(4 - x - b)}^{2} = R^{2}

\Leftrightarrow {2 x}^{2} - 2 (a - b + 4) x + a^{2} + {(b - 4)}^{2} - R^{2} = 0

has unique root .

\Leftrightarrow Δ^{'} = {(a - b + 4)}^{2} - {2 a}^{2} - 2 {(b - 4)}^{2} + {2 R}^{2} = 0

\Leftrightarrow - a^{2} + 2 (4 - b) a - {(4 - b)}^{2} + {2 R}^{2} = 0 . From (*) we get :

\Leftrightarrow - a^{2} + 8 a - 2 ab - b^{2} + 8 b - 16 + {(a - b)}^{2} = 0

\Leftrightarrow 4 (2 - b) a + 8 b - 16 = 0

\Leftrightarrow (2 - b) a + 2 b - 4 = 0 (3)

i) If b = 2 then (3) is true \Rightarrow R = 2, a = 2 \pm 2 \sqrt{2}

ii) If b \neq 2 then (3) \Leftrightarrow a = 2

Replace into (*) we get :

R = \frac{2}{(1 + \sqrt{2})} = 2 (\sqrt{2} - 1) = b, or R = \frac{2}{\sqrt{2} - 1} = 2 (\sqrt{2} + 1) = - b