a-circle-is-tangent-to-x-axis-y-axis-and-the-line-3x-4y-6-0-what-its-the-equation-

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 \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 :

{\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 \dots 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