Solve Lesson 54 Page 89 SBT Math 10 – Kite>

[ad_1]

Topic

Write the equation of the circle (OLD) in each of the following cases:

a) (OLD) has a mind I(−6 ; 2) radius 7

b) (OLD) has a mind I(3 ; – 7) and passes through point A(4 ; 1)

c) (OLD) has a mind I(1 ; 2) and tangent to the line 3x + 4y + 19 = 0

d) (OLD) has a diameter AB with A(−2 ; 3) and REMOVE(0 ; 1)

e) (OLD) has a mind I belongs to the line \({\Delta _1}:\left\{ \begin{array}{l}x = 1 + t\\y = 1 – t\end{array} \right.\) and (OLD) tangent to two lines2: 3x + 4y – 1 = 0,3: 3x – 4y + 2 = 0

Solution method – See details

+) From a and d determine the radius of (C) and then write PT of the circle in canonical form

+) Consider sentence e

Step 1: Parameterizing center coordinates I

Step 2: Create a PT from the hypothesis: \(d(I,{\Delta _2}) = d(I,{\Delta _3})\)

Step 3: Solve the PT found in step 2 to find the coordinates of the center I and the radius of the circle and then write PT of the circle in canonical form

}

Detailed explanation

a) (OLD) has a mind I(−6 ; 2) radius 7 should have PT: \({(x + 6)^2} + {(y – 2)^2} = 49\)

b) (OLD) has a mind I(3 ; – 7) and passes through point A(4 ; 1) \( \Rightarrow \) Radius of (OLD) is \(IA = \sqrt {{{(4 – 3)}^2} + {{(1 + 7)}^2}} = \sqrt {65} \)

\( \Rightarrow \)(OLD) has PT: \({(x – 3)^2} + {(y + 7)^2} = 65\)

c) (OLD) has a mind I(1 ; 2) and tangent to the line 3x + 4y + 19 = 0

\( \Rightarrow \) Radius of (OLD) is the distance from the center I to the straight line : 3x + 4y + 19 = 0

We have: \(d(I,\Delta ) = \frac{{\left| {3.1 + 4.2 + 19} \right|}}{{\sqrt {{3^2} + {4^2}} } } = \frac{{30}}{5} = 6\)

\( \Rightarrow \)(OLD) has PT: \({(x – 1)^2} + {(y – 2)^2} = 36\)

d) (OLD) has a diameter AB with A(−2 ; 3) and REMOVE(0 ; 1)

\( \Rightarrow \) (OLD) has a mind I is the midpoint of AB \( \Rightarrow I( – 1,2)\)

(OLD) has a radius IA = IB = \(\sqrt 2 \)

\( \Rightarrow \)(OLD) has PT: \({(x + 1)^2} + {(y – 2)^2} = 2\)

e) (OLD) has a mind I belongs to the line \({\Delta _1}:\left\{ \begin{array}{l}x = 1 + t\\y = 1 – t\end{array} \right.\) and (OLD) tangent to two lines2: 3x + 4y – 1 = 0,3: 3x – 4y + 2 = 0

Since \(I \in {\Delta _1}\) so \(I(1 + t;1 – t)\)

By assumption, \(R = d(I,{\Delta _2}) = d(I,{\Delta _3}) \Leftrightarrow \frac{{\left| {3(1 + t) + 4(1 – t) – 1} \right|}}{{\sqrt {{3^2} + {4^2}} }} = \frac{{\left| {3(1 + t) – 4(1 – t) ) + 2} \right|}}{{\sqrt {{3^2} + {{( – 4)}^2}} }}\)

\( \Leftrightarrow \left| {6 – t} \right| = \left| {7t + 1} \right| \Leftrightarrow \left[\begin{array}{l}6–t=7t+1\\6–t=–7t–1\end{array}\right\Leftrightarrow\left[\begin{array}{l}t=\frac{5}{8}\\t=\frac{{–7}}{6}\end{array}\right\)[\begin{array}{l}6–t=7t+1\\6–t= –7t–1\end{array}\right\Leftrightarrow\left[\begin{array}{l}t=\frac{5}{8}\\t=\frac{{–7}}{6}\end{array}\right\)

With \(t = \frac{5}{8} \Rightarrow I\left( {\frac{{13}}{8};\frac{3}{8}} \right)\) \( \Rightarrow \ )\(R = \frac{{43}}{{40}}\). Then (OLD) has PT: \({\left( {x – \frac{{13}}{8}} \right)^2} + {\left( {y – \frac{3}{8}} \right) ^2} = \frac{{1849}}{{1600}}\)

With \(t = – \frac{7}{6} \Rightarrow I\left( { – \frac{1}{6};\frac{{13}}{6}} \right)\)\( \ Rightarrow \)\(R = \frac{{43}}{{30}}\). Then (OLD) has PT: \({\left( {x + \frac{1}{6}} \right)^2} + {\left( {y – \frac{{13}}{6}} \right) ^2} = \frac{{1849}}{{900}}\)

[ad_2]

Source link net do edu

Leave a Reply