## Solve Exercises Lesson 21: Circle in Coordinate Plane (Connection) – Math Book

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) x2 + y2 + xy + 4x – 2 = 0

b) x2 + y2 – 2y – 4x + 5 = 0

c) x2 + y2 + 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) x2 + y2 + xy + 4x – 2 = 0

b) x2 + y2 – 2y – 4x + 5 = 0

c) x2 + y2 + 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) x2 + y2 + 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) x2 + y2 – 2y – 4x + 5 = 0

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

Consider: a2 + b2 – c = 0

The equation above is not the equation of a circle.

c) x2 + y2 + 6x – 8y + 1 =0

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

Consider: a2 + b2 – 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.

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): x2 + y2 + 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): x2 + y2 + 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 02 + 22 + 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 to; 4 + costo).

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 to; 4 + costo).

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 0o; 4 + cos 0o) = (2; 5)

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

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

We have: x = 2 + sin to and y = 4 + costo

=> x – 2 = sin to and y – 4 = costo

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.