a circle is tangent to x−axis , y−axis and the line 3x−4y+6=0. what its the equation?


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Question Number 116239 by bemath last updated on 02/Oct/20

$a circle is tangent to x - axis, y - axis$ $and the line 3 x - 4 y + 6 = 0 .$ $what its the equation ?$
	Answered by bemath last updated on 02/Oct/20
	$say P(a,b) be a centre point the circle tangent to x−axis →radius = ∣b∣ tangent to y−axis→radius = ∣a∣ tangent to the line 3x−4y+6=0 radius = ((∣3a−4b+6∣)/5) Now we have ∣a∣ = ∣b∣ → { ((a=b)),((a=−b)) :} for case a=b ⇒∣a∣ = ((∣6−a∣)/5) ⇒ 25a^2 = a^2 −12a+36 ⇒24a^2 +12a−36=0 ⇒2a^2 +a−3=0 ,(2a+3)(a−1)=0 → { ((a=−(3/2)=b⇒→(x+(3/2))^2 +(y+(3/2))^2 =(9/4))),((a=1=b⇒(x−1)^2 +(y−1)^2 =1)) :} for case a=−b ⇒∣a∣ = ((∣7a+6∣)/5) ⇒25a^2 = 49a^2 +84a+36 ⇒24a^2 +84a+36=0 ⇒2a^2 +7a+3=0,(2a+1)(a+3)=0 → { ((a=−(1/2),b=(1/2)⇒(x+(1/2))^2 +(y−(1/2))^2 =(1/4))),((a=−3,b=3⇒(x+3)^2 +(y−3)^2 =9)) :}$
	$say P (a, b) be a centre point the circle$ $tangent to x - axis \to radius = ∣ b ∣$ $tangent to y - axis \to radius = ∣ a ∣$ $tangent to the line 3 x - 4 y + 6 = 0$ $radius = \frac{∣ 3 a - 4 b + 6 ∣}{5}$ $Now we have ∣ a ∣ = ∣ b ∣ \to {\begin{cases} a = b \\ a = - b \end{cases}$ $for case a = b \Rightarrow∣ a ∣ = \frac{∣ 6 - a ∣}{5}$ $\Rightarrow {25 a}^{2} = a^{2} - 12 a + 36$ $\Rightarrow {24 a}^{2} + 12 a - 36 = 0$ $\Rightarrow {2 a}^{2} + a - 3 = 0, (2 a + 3) (a - 1) = 0$ $\to {\begin{cases} a = - \frac{3}{2} = b \Rightarrow\to {(x + \frac{3}{2})}^{2} + {(y + \frac{3}{2})}^{2} = \frac{9}{4} \\ a = 1 = b \Rightarrow {(x - 1)}^{2} + {(y - 1)}^{2} = 1 \end{cases}$ $for case a = - b$ $\Rightarrow∣ a ∣ = \frac{∣ 7 a + 6 ∣}{5}$ $\Rightarrow {25 a}^{2} = {49 a}^{2} + 84 a + 36$ $\Rightarrow {24 a}^{2} + 84 a + 36 = 0$ $\Rightarrow {2 a}^{2} + 7 a + 3 = 0, (2 a + 1) (a + 3) = 0$ $\to {\begin{cases} a = - \frac{1}{2}, b = \frac{1}{2} \Rightarrow {(x + \frac{1}{2})}^{2} + {(y - \frac{1}{2})}^{2} = \frac{1}{4} \\ a = - 3, b = 3 \Rightarrow {(x + 3)}^{2} + {(y - 3)}^{2} = 9 \end{cases}$
	Commented by bobhans last updated on 02/Oct/20

	Answered by 1549442205PVT last updated on 02/Oct/20
	$Suppose the equation of the circle is (x−a)^2 +(y−b)^2 =R^2 From the hypothesis we infer the system of equations : { (((x−a)^2 +(y−b)^2 =R^2 )),((x=0)) :}(1) { (((x−a)^2 +(y−b)^2 =R^2 )),((y=0)) :}(2) { (((x−a)^2 +(y−b)^2 =R^2 )),((3x−4y+6=0)) :}(3) have unique root (1)⇔a^2 +y^2 −2yb+b^2 −R^2 =0 with Δ′=b^2 −a^2 −b^2 +R^2 =R^2 −a^2 (2)⇔x^2 −2ax+a^2 +b^2 −R^2 =0 with Δ′=a^2 −a^2 −b^2 +R^2 =R^2 −b^2 (3)⇔x^2 −2ax+a^2 +(((3x+6)/4)−b)^2 −R^2 =0 ⇔x^2 −2ax+a^2 +((9x^2 +36x+36)/(16))−(((3x+6)b)/2)+b^2 −R^2 =0 ⇔25x^2 −(32a+24b−36)x+16(a^2 +b^2 )−48b+36−16R^2 =0 with Δ′=(16a+12b−18)^2 −25[16(a^2 +b^2 )−48b+36−16R^2 ] =−144a^2 −256b^2 −576a+384ab +768b−576+400R^2 (4) We need must have Δ′=0 so { ((−a^2 +R^2 =0)),((R^2 −b^2 =0)) :}⇔ { ((a=±R)),((b=±R)) :}(5) i)For a=R,b=R replace into (4)we get −144R^2 −256R^2 −576R+384R^2 +768R−586+400R^2 =0 ⇔384R^2 +192R−576=0 ⇔2R^2 +R−3=0⇒R=1=a=b (x−1)^2 +(y−1)^2 =1 ii)For a=R,b=−R we get −144R^2 −256R^2 −576R−384R^2 −768R −576+400R^2 ⇔384R^2 +1344R+576=0 2R^2 +7R+3=0⇒has no roots iii)For a=−R,b=R we get: −144R^2 −256R^2 +576R−384R^2 +768R−576+400R^2 =0 ⇔384R^2 −1344R+576=0 ⇔2R^2 −7R+3=0 R=3 ∨(1/2) we have two the circles (x+3)^2 +(y−3)^2 =9 (R=3=b=−a) (x+(1/2))^2 +(y−(1/2))^2 =(1/4) (R=(1/2)=b=−a) iv)For a=−R,b=−R we get −144R^2 −256R^2 +576R+384R^2 −768R−576+400R^2 =0 ⇔384R^2 −192R−576=0 ⇔2R^2 −R−3=0⇒R=(3/2) =−a=−b (x+(3/2))^2 +(y+(3/2))^2 =(9/4) Thus,we get four circle with the equations as above$
	$Suppose the equation of the circle is$ ${(x - a)}^{2} + {(y - b)}^{2} = R^{2}$ $From the hypothesis we infer the system$ $of equations :$ ${\begin{cases} {(x - a)}^{2} + {(y - b)}^{2} = R^{2} \\ x = 0 \end{cases} (1)$ ${\begin{cases} {(x - a)}^{2} + {(y - b)}^{2} = R^{2} \\ y = 0 \end{cases} (2)$ ${\begin{cases} {(x - a)}^{2} + {(y - b)}^{2} = R^{2} \\ 3 x - 4 y + 6 = 0 \end{cases} (3)$ $have unique root$ $(1) \Leftrightarrow a^{2} + y^{2} - 2 yb + b^{2} - R^{2} = 0$ $with Δ^{'} = b^{2} - a^{2} - b^{2} + R^{2} = R^{2} - a^{2}$ $(2) \Leftrightarrow x^{2} - 2 ax + a^{2} + b^{2} - R^{2} = 0$ $with Δ^{'} = a^{2} - a^{2} - b^{2} + R^{2} = R^{2} - b^{2}$ $(3) \Leftrightarrow x^{2} - 2 ax + a^{2} + {(\frac{3 x + 6}{4} - b)}^{2} - R^{2} = 0$ $\Leftrightarrow x^{2} - 2 ax + a^{2} + \frac{{9 x}^{2} + 36 x + 36}{16} - \frac{(3 x + 6) b}{2} + b^{2} - R^{2} = 0$ $\Leftrightarrow {25 x}^{2} - (32 a + 24 b - 36) x + 16 (a^{2} + b^{2}) - 48 b + 36 - {16 R}^{2} = 0$ $with Δ^{'} = {(16 a + 12 b - 18)}^{2} - 25 [16 (a^{2} + b^{2}) - 48 b + 36 - {16 R}^{2}]$ $= - {144 a}^{2} - {256 b}^{2} - 576 a + 384 ab$ $+ 768 b - 576 + {400 R}^{2} (4)$ $We need must have Δ^{'} = 0 so$ ${\begin{cases} - a^{2} + R^{2} = 0 \\ R^{2} - b^{2} = 0 \end{cases} \Leftrightarrow {\begin{cases} a = \pm R \\ b = \pm R \end{cases} (5)$ $i) For a = R, b = R replace into (4) we get$ $- {144 R}^{2} - {256 R}^{2} - 576 R + {384 R}^{2}$ $+ 768 R - 586 + {400 R}^{2} = 0$ $\Leftrightarrow {384 R}^{2} + 192 R - 576 = 0$ $\Leftrightarrow {2 R}^{2} + R - 3 = 0 \Rightarrow R = 1 = a = b$ ${(x - 1)}^{2} + {(y - 1)}^{2} = 1$ $ii) For a = R, b = - R we get$ $- {144 R}^{2} - {256 R}^{2} - 576 R - {384 R}^{2} - 768 R$ $- 576 + {400 R}^{2} \Leftrightarrow {384 R}^{2} + 1344 R + 576 = 0$ ${2 R}^{2} + 7 R + 3 = 0 \Rightarrow has no roots$ $iii) For a = - R, b = R we get :$ $- {144 R}^{2} - {256 R}^{2} + 576 R - {384 R}^{2}$ $+ 768 R - 576 + {400 R}^{2} = 0$ $\Leftrightarrow {384 R}^{2} - 1344 R + 576 = 0$ $\Leftrightarrow {2 R}^{2} - 7 R + 3 = 0$ $R = 3 \lor \frac{1}{2} we have two the circles$ ${(x + 3)}^{2} + {(y - 3)}^{2} = 9 (R = 3 = b = - a)$ ${(x + \frac{1}{2})}^{2} + {(y - \frac{1}{2})}^{2} = \frac{1}{4} (R = \frac{1}{2} = b = - a)$ $iv) For a = - R, b = - R we get$ $- {144 R}^{2} - {256 R}^{2} + 576 R + {384 R}^{2}$ $- 768 R - 576 + {400 R}^{2} = 0$ $\Leftrightarrow {384 R}^{2} - 192 R - 576 = 0$ $\Leftrightarrow {2 R}^{2} - R - 3 = 0 \Rightarrow R = \frac{3}{2} = - a = - b$ ${(x + \frac{3}{2})}^{2} + {(y + \frac{3}{2})}^{2} = \frac{9}{4}$ $Thus, we get four circle with the$ $equations as above$
	Commented by bemath last updated on 02/Oct/20

	$thank you$
	Answered by TANMAY PANACEA last updated on 02/Oct/20

	$e q n o f c i r c l e {(x - α)}^{2} + {(y - α)}^{2} = α^{2}$ $∣ \frac{3 α - 4 α + 6}{\sqrt{3^{2} + {(- 4)}^{2}}} ∣= α$ $36 - 12 α + α^{2} = 25 α^{2}$ $12 (2 α^{2} + α - 3) = 0$ $2 α^{2} + 3 α - 2 α - 3 =$ $α (2 α + 3) - 1 (2 α + 3) = 0$ $(2 α + 3) (α - 1) = 0$ $e q n c i r c l e$ ${(x - 1)}^{2} + {(y - 1)}^{2} = 1$ ${(x + \frac{3}{2})}^{2} + {(y + \frac{3}{2})}^{2} = {(\frac{3}{2})}^{2}$ $o k . . . i a m c o m p l e t i n g$ $x^{2} + y^{2} - 2 x - 2 y + 2 = 1$ $x^{2} + y^{2} - 2 x - 2 y + 1 = 0 \to f i r s t e q n$ $x^{2} + y^{2} + 3 x + 3 y + \frac{9}{4} + \frac{9}{4} = \frac{9}{4}$ $x^{2} + y^{2} + 3 x + 3 y + 2.25 = 0$
	Commented by bemath last updated on 02/Oct/20

	$not complete sir ?$
	Commented by bemath last updated on 02/Oct/20

	$i^{'} m got 4 equation of a circle . it right ?$
	Answered by 1549442205PVT last updated on 02/Oct/20

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