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**Solution for Exercise 21: Circles in the coordinate plane (Connection)**

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### Solve problem 7.13 page 46 Math textbook 10 Connecting knowledge volume 2

Find the center and radius of the circle: (x + 3)^{2} + (y – 3)^{2} = 36

Solve problem 7.13 page 46 Math textbook 10 Connecting knowledge volume 2

Find the center and radius of the circle: (x + 3)^{2} + (y – 3)^{2} = 36

**Solution method**

The equation of the circle (C): \({\left( {x – a} \right)^2} + {\left( {y – b} \right)^2} = {R^2}\)

There is a point \(M\left( {x;y} \right)\) on the circle (C), center ((a; b), radius R

**Detailed explanation**

The circle has center I(-3; 3) and radius R = \(\sqrt{36}=6\).

### Solve problem 7.14 page 46 Math textbook 10 Connecting knowledge volume 2

Indicate which of the following equations is an equation of a circle and find the center and radius of the circle respectively.

a) x^{2 }+ y^{2} + xy + 4x – 2 = 0

b) x^{2 }+ y^{2} – 2y – 4x + 5 = 0

c) x^{2 }+ y^{2} + 6x – 8y + 1 =0

Solve problem 7.14 page 46 Math textbook 10 Connecting knowledge volume 2

Indicate which of the following equations is an equation of a circle and find the center and radius of the circle respectively.

a) x^{2 }+ y^{2} + xy + 4x – 2 = 0

b) x^{2 }+ y^{2} – 2y – 4x + 5 = 0

c) x^{2 }+ y^{2} + 6x – 8y + 1 =0

**Solution method**

The equation \({x^2} + {y^2} – 2{\rm{a}}x – 2by + c = 0\) is the equation of a circle (C) if and only if \({ a^2} + {b^2} – c > 0\). Then, (C) has center I(a; b) and radius \(R = \sqrt {{a^2} + {b^2} – c} \)

**Detailed explanation**

a) x^{2 }+ y^{2} + xy + 4x – 2 = 0 is not a circle equation because it does not conform to the general form of the equation of a circle.

b) x^{2 }+ y^{2} – 2y – 4x + 5 = 0

We have: a = 1, b = 2, c = 5

Consider: a^{2} + b^{2} – c = 0

The equation above is not the equation of a circle.

c) x^{2 }+ y^{2} + 6x – 8y + 1 =0

We have: a = -3, b = 4, c = 1

Consider: a^{2} + b^{2} – c = 24 > 0.

It follows that the above equation is the equation of a circle, with center I(-3; 4) and radius R = \(\sqrt{24}\)

### Solve problem 7.15 on page 47 Math textbook 10 Connecting knowledge volume 2

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

a) Has center I(-2; 5) and radius R = 7.

b) Has center I(1; -2) and passes through point A(-2; 2)

c) Has a diameter AB, with A(-1; -3), B(-3; 5)

d) Having center I(1; 3) and tangent to the line x + 2y + 3 = 0.

Solve problem 7.15 on page 47 Math textbook 10 Connecting knowledge volume 2

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

a) Has center I(-2; 5) and radius R = 7.

b) Has center I(1; -2) and passes through point A(-2; 2)

c) Has a diameter AB, with A(-1; -3), B(-3; 5)

d) Having center I(1; 3) and tangent to the line x + 2y + 3 = 0.

**Solution method**

The equation \({x^2} + {y^2} – 2{\rm{a}}x – 2by + c = 0\) is the equation of a circle (C) if and only if \({ a^2} + {b^2} – c > 0\). Then, (C) has center I(a; b) and radius \(R = \sqrt {{a^2} + {b^2} – c} \)

**Detailed explanation**

a) The equation of the circle is: (x +2)^{2} + (y -5)^{2} = 49.

b) Circle with radius R = IA = \(\sqrt{(1+2)^{2}+(-2-2)^{2}}=5\)

The equation of the circle is: (x – 1)^{2} + (y + 2)^{2} = 25.

c)

+ Diameter of circle: AB = \(\sqrt{(-3+1)^{2}+(5+3)^{2}}=\sqrt{68}\)

Derive a circle with radius R = \(\frac{AB}{2}=\sqrt{17}\)

+ The center of the circle is midpoint I of line segment AB, so I \(\left ( \frac{-1-3}{2};\frac{-3+5}{2} \right )=(- 2;1)\)

The equation of the circle is: (x + 2)^{2} + (y – 1)^{2} = 17.

d) The circle is tangent to the line (d): x + 2y + 3 = 0, so the radius of the circle is equal to the distance from range I to the line.

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We have: \(d_{(I;d)}=\frac{|1+2.3+3|}{\sqrt{1^{2}+2^{2}}}=2\sqrt{5}\ ) = R.

The equation of the circle is: (x – 1)^{2} + (y – 3)^{2} = 20.

### Solve problem 7.16 page 47 Math textbook 10 Connecting knowledge volume 2

In the coordinate plane, let ABC be a triangle with A(6; -2), B(4; 2), C(5; -5). Write the equation of the circle circumscribing the triangle.

Solve problem 7.16 page 47 Math textbook 10 Connecting knowledge volume 2

In the coordinate plane, let ABC be a triangle with A(6; -2), B(4; 2), C(5; -5). Write the equation of the circle circumscribing the triangle.

**Solution method**

We have:

\(IA=\sqrt{(x-6)^{2}+(y+2)^{2}}\),

\(IB= \sqrt{(x-4)^{2}+(y-2)^{2}}\),

\(IC= \sqrt{(x-5)^{2}+(y+5)^{2}}\)

With IC = IA = IB, consider the system of equations \(\left\{ {\begin{array}{*{20}{c}}

{{{(x – 6)}^2} + {{(y + 2)}^2} = {{(x – 4)}^2} + {{(y – 2)}^2}}\ \

{{{(x – 4)}^2} + {{(y – 2)}^2} = {{(x – 5)}^2} + {{(y + 5)}^2}}

\end{array}} \right.\). Solve the system => center of the circle

Calculate IA, then deduce the equation of the circle to find

**Detailed explanation**

Call the circumcircle of triangle ABC with center I(x; y).

Since I is the center of the circumcircle of triangle ABC, so I is equidistant from 3 vertices A, B, C. Or IA = IB = IC

\(IA=\sqrt{(x-6)^{2}+(y+2)^{2}}\),

\(IB= \sqrt{(x-4)^{2}+(y-2)^{2}}\),

\(IC= \sqrt{(x-5)^{2}+(y+5)^{2}}\)

Since IC = IA = IB, we have a system of equations:

\(\left\{\begin{matrix}(x-6)^{2}+(y+2)^{2}=(x-4)^{2}+(y-2)^{2} \\ (x-4)^{2}+(y-2)^{2}=(x-5)^{2}+(y+5)^{2}\end{matrix}\right.\ \\Leftrightarrow \left\{\begin{matrix}-12x+36+4y+4=-8x+16-4y+4\\ -8x+16-4y+4=-10x+25+10y+25\end {matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix}x=1\\ y=-2\end{matrix}\right.\)

=> circle with center I(1; -2)

+) Calculate IA = \(\sqrt{(1-6)^{2}+(-2+2)^{2}}\) = 5

So the equation of the circle is: (x – 1)^{2} + (y+2)^{2} = 25.

### Solve problem 7.17, page 47 Math textbook 10 Connecting knowledge volume 2

Given circle (C): x^{2} + y^{2} + 2x – 4y + 4 = 0. Write an equation for the tangent line d to (C) at the point M(0; 2).

Solve problem 7.17, page 47 Math textbook 10 Connecting knowledge volume 2

Given circle (C): x^{2} + y^{2} + 2x – 4y + 4 = 0. Write an equation for the tangent line d to (C) at the point M(0; 2).

**Solution method**

Determine the center I of the circle (C) and the normal vector. Then write down the required equation

**Detailed explanation**

+) Do 0^{2} + 2^{2} + 2.0 – 4.2 + 4 = 0, so M belongs to the circle (C).

+) The circle (C) has center I(-1; 2). The tangent to (C) at M has a normal vector \(\overrightarrow{IM}(1; 0)\), so the equation is:

1(x – 0) + 0.(y – 2) = 0 or x =0.

### Solve problem 7.18 page 47 Math textbook 10 Connecting knowledge volume 2

The motion of an object over a period of 180 minutes is represented in the coordinate plane. Accordingly, at time t ($0\leq t\leq 180$) the object is at a position with coordinates (2 + sin t^{o}; 4 + cost^{o}).

a) Find the starting and ending positions of the object.

b) Find the motion of the object.

Solve problem 7.18 page 47 Math textbook 10 Connecting knowledge volume 2

The motion of an object over a period of 180 minutes is represented in the coordinate plane. Accordingly, at time t ($0\leq t\leq 180$) the object is at a position with coordinates (2 + sin t^{o}; 4 + cost^{o}).

a) Find the starting and ending positions of the object.

b) Find the motion of the object.

**Solution method**

Determine the coordinates of the point at the initial position of the object at time t = 0

Determine the coordinates of the point at the end position of the object at time t = 180

**Detailed explanation**

a) The initial position of the object is at time t = 0, so the coordinates of the point are: (2 + sin 0^{o}; 4 + cos 0^{o}) = (2; 5)

The end position of the object is at time t = 180, so the coordinates of the point are: (2 + sin 180^{o}; 4 + cos 180^{o}) = (2; 3)

b) Call the point M(x; y) belonging to the motion of the body.

We have: x = 2 + sin t^{o }and y = 4 + cost^{o}

=> x – 2 = sin t^{o} and y – 4 = cost^{o}

Which \(sin^{2}t^{o}+cos^{2}t^{o}=1\)

So (x – 2)^{2} + (y – 4)^{2} =1

So the motion trajectory of the object is a circle with center I(2; 4) and radius equal to 1.