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.


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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
	$let P(2,b) be a centre point a circle touches x−axis ; line y=x and y+x=4 (i) b = ((∣2−b∣)/( (√2))) ⇒2b^2 =4−4b+b^2 b^2 +4b−4=0; (b+2)^2 =8 b = 2(√2)−2 (ii) b = ((∣2+b−4∣)/( (√2))) ⇒b=2(√2) −2 so we have → { ((P(2, 2(√2) −2))),((r = 2(√2)−2)) :} equation of a circle is (x−2)^2 +(y−2(√2)+2)^2 =(2(√2)−2)^2 (x−2)^2 +(y−2(√2)+2)^2 =12−8(√2)$
	$l e t P (2, b) b e a c e n t r e p o i n t a c i r c l e$ $t o u c h e s x - a x i s; l i n e y = x a n d y + x = 4$ $(i) b = \frac{∣ 2 - b ∣}{\sqrt{2}} \Rightarrow 2 b^{2} = 4 - 4 b + b^{2}$ $b^{2} + 4 b - 4 = 0; {(b + 2)}^{2} = 8$ $b = 2 \sqrt{2} - 2$ $(i i) b = \frac{∣ 2 + b - 4 ∣}{\sqrt{2}} \Rightarrow b = 2 \sqrt{2} - 2$ $s o w e h a v e \to {\begin{cases} P (2, 2 \sqrt{2} - 2) \\ r = 2 \sqrt{2} - 2 \end{cases}$ $e q u a t i o n o f a c i r c l e i s$ ${(x - 2)}^{2} + {(y - 2 \sqrt{2} + 2)}^{2} = {(2 \sqrt{2} - 2)}^{2}$ ${(x - 2)}^{2} + {(y - 2 \sqrt{2} + 2)}^{2} = 12 - 8 \sqrt{2}$
	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
	$a)Suppose the equation of the circle C is (x−a)^2 +(y−b)^2 =R^2 .Since C touches the lines y=0 and y=x ,we infer the equations (x−a)^2 +b^2 =R^2 (1) and (x−a)^2 +(x−b)^2 =R^2 (2)has unique root (1)⇔x^2 −2ax+a^2 +b^2 −R^2 =0 with Δ′=a^2 −a^2 −b^2 +R^2 =R^2 −b^2 (2)⇔2x^2 −2(a+b)x+a^2 +b^2 −R^2 =0 with Δ′=(a+b)^2 −2a^2 −2b^2 +2R^2 =2R^2 −(a−b)^2 .Therefore,(1)(2)has unique root if and only if { ((R^2 −b^2 =0)),((2R^2 −(a−b)^2 =0)) :}⇔ { ((b=±R)),((a=±R±R(√2))) :} (∗) Hence,the equations of C are:(four cases) i)b=R⇒a=R(1±(√2))gives two the circles family: (x−R(1+(√2)))^2 +(y−R)^2 =R^2 . (x−R(1−(√2)))^2 +(y−R)^2 =R^2 ii)b=−R⇒a=R(−1±(√2))gives two the circles family: [x−R(−1+(√2))]^2 +(y+R)^2 =R^2 [(x−R(−1−(√2))]^2 +(y+R)^2 =R^2 Example,R=1,b=1,a=1+(√2) .... b)Now C touches to the x+y=4 if and only if the equation (x−a)^2 +(4−x−b)^2 =R^2 ⇔2x^2 −2(a−b+4)x+a^2 +(b−4)^2 −R^2 =0 has unique root. ⇔Δ′=(a−b+4)^2 −2a^2 −2(b−4)^2 +2R^2 =0 ⇔−a^2 +2(4−b)a−(4−b)^2 +2R^2 =0.From(∗)we get: ⇔−a^2 +8a−2ab−b^2 +8b−16+(a−b)^2 =0 ⇔4(2−b)a+8b−16=0 ⇔(2−b)a+2b−4=0(3) i)If b=2 then (3) is true⇒ R=2,a=2±2(√2) ii)If b≠2 then (3)⇔a=2 Replace into (∗)we get: R=(2/((1+(√2))))=2((√2)−1)=b,or R=(2/( (√2)−1))=2((√2)+1)=−b$
	$a) Suppose the equation of the circle C is$ ${(x - a)}^{2} + {(y - b)}^{2} = R^{2} . Since C touches$ $the lines y = 0 and y = x, we infer the$ $equations {(x - a)}^{2} + b^{2} = R^{2} (1)$ $and {(x - a)}^{2} + {(x - b)}^{2} = R^{2} (2) has unique$ $root$ $(1) \Leftrightarrow x^{2} - 2 ax + a^{2} + b^{2} - R^{2} = 0$ $with Δ^{'} = a^{2} - a^{2} - b^{2} + R^{2} = R^{2} - b^{2}$ $(2) \Leftrightarrow {2 x}^{2} - 2 (a + b) x + a^{2} + b^{2} - R^{2} = 0$ $with Δ^{'} = {(a + b)}^{2} - {2 a}^{2} - {2 b}^{2} + {2 R}^{2}$ $= {2 R}^{2} - {(a - b)}^{2} . Therefore, (1) (2) has$ $unique root if and only if$ ${\begin{cases} R^{2} - b^{2} = 0 \\ {2 R}^{2} - {(a - b)}^{2} = 0 \end{cases} \Leftrightarrow {\begin{cases} b = \pm R \\ a = \pm R \pm R \sqrt{2} \end{cases} ()$ $Hence, the equations of C are : (four cases)$ $i) b = R \Rightarrow a = R (1 \pm \sqrt{2}) gives two the circles family :$ ${(x - R (1 + \sqrt{2}))}^{2} + {(y - R)}^{2} = R^{2} .$ ${(x - R (1 - \sqrt{2}))}^{2} + {(y - R)}^{2} = R^{2}$ $ii) b = - R \Rightarrow a = R (- 1 \pm \sqrt{2}) gives two the circles family :$ ${[x - R (- 1 + \sqrt{2})]}^{2} + {(y + R)}^{2} = R^{2}$ $[{(x - R (- 1 - \sqrt{2})]}^{2} + {(y + R)}^{2} = R^{2}$ $Example, R = 1, b = 1, a = 1 + \sqrt{2} . . . .$ $b) Now C touches to the x + y = 4 if and$ $only if the equation {(x - a)}^{2} + {(4 - x - b)}^{2} = R^{2}$ $\Leftrightarrow {2 x}^{2} - 2 (a - b + 4) x + a^{2} + {(b - 4)}^{2} - R^{2} = 0$ $has unique root .$ $\Leftrightarrow Δ^{'} = {(a - b + 4)}^{2} - {2 a}^{2} - 2 {(b - 4)}^{2} + {2 R}^{2} = 0$ $\Leftrightarrow - a^{2} + 2 (4 - b) a - {(4 - b)}^{2} + {2 R}^{2} = 0 . From () we get :$ $\Leftrightarrow - a^{2} + 8 a - 2 ab - b^{2} + 8 b - 16 + {(a - b)}^{2} = 0$ $\Leftrightarrow 4 (2 - b) a + 8 b - 16 = 0$ $\Leftrightarrow (2 - b) a + 2 b - 4 = 0 (3)$ $i) If b = 2 then (3) is true \Rightarrow R = 2, a = 2 \pm 2 \sqrt{2}$ $ii) If b \neq 2 then (3) \Leftrightarrow a = 2$ $Replace into (*) we get :$ $R = \frac{2}{(1 + \sqrt{2})} = 2 (\sqrt{2} - 1) = b, or R = \frac{2}{\sqrt{2} - 1} = 2 (\sqrt{2} + 1) = - b$
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