## Solving Exercise 5: Equation of a circle (C7 – Math 10 Kite) – Math Book

Solution of Exercise 5: Equation of a circle (C7 – Math 10 Kite)
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### Solve the exercises Lesson 1, page 91, Math Textbook 10, Kite episode 2

Which of the following is an equation of a circle?

a) $${x^2} + {y^2} – 2x + 2y – 7 = 0$$

b) $${x^2} + {y^2} – 8x + 2y + 20 = 0$$

Solution method

The equation $${x^2} + {y^2} – 2{\rm{a}}x – 2by + c = 0$$ is a circle equation if and only if $${a^2} + {b^2} – c > 0$$

Solution guide

a) Since $${1^2} + {\left( { – 1} \right)^2} > – 7$$ so $${x^2} + {y^2} – 2x + 2y – 7 = 0$$ is the equation of the circle

b) Since $${4^2} + {\left( { – 1} \right)^2} < 20$$ $${x^2} + {y^2} – 8x + 2y + 20 = 0$$ is not the equation of the circle

### Solve the exercises Lesson 2 page 91 Math textbook 10 Kite episode 2

Find the center and radius of the circle in the following case:

a) The circle has the equation$${(x + 1)^2} + {(y – 5)^2} = 9$$ ;

b) The circle has the equation$${x^2} + {y^2}-6x – 2y-{\rm{1}}5 = 0$$ .

Solution method

a) The equation $${\left( {x – a} \right)^2} + {\left( {y – b} \right)^2} = {R^2}$$ centered at $$I\left( {a;b} \right)$$ and radius R

b) The equation $${x^2} + {y^2} – 2{\rm{a}}x – 2by + c = 0$$ has center $$I\left( {a;b} \right) )$$ and radius $$R = \sqrt {{a^2} + {b^2} – c}$$

Solution guide

a) The circle $${(x + 1)^2} + {(y – 5)^2} = 9$$ has center $$I\left( { – 1;5} \right)$$ and \ (R = 3\)

b) The circle $${x^2} + {y^2}-6x – 2y-{\rm{1}}5 = 0$$ has center $$I\left( {3;1} \right)$$ and $$R = \sqrt {{3^2} + {1^2} + 15} = 5$$

### Solve the exercises Lesson 3 page 91 Math textbook 10 Kite episode 2

Write an equation for the circle in each of the following cases:

a) A circle with center I(- 3 ; 4) radius R = 9;

b) The circle has center I(5 ;-2) and passes through the point M(4;- 1);

c) The circle has center I(1;- 1) and has a tangent A: 5x- 12y – 1 = 0;

d) A circle with diameter AB with A(3;-4) and B(-1; 6);

e) The circle passes through three points A(1;1), B(3; 1), C(0; 4).

Solution method

The circle with center $$I\left( {a;b} \right)$$ and radius R has the equation: $${\left( {x – a} \right)^2} + {\left ( {y – b} \right)^2} = {R^2}$$

Solution guide

a) The equation of the circle is: $${\left( {x + 3} \right)^2} + {\left( {y – 4} \right)^2} = 81$$

b) The radius of the circle is: $$R = IM = \sqrt {{{\left( {4 – 5} \right)}^2} + {{\left( { – 1 + 2} \right)} ^2}} = \sqrt 2$$

The equation of the circle is: $${\left( {x – 5} \right)^2} + {\left( {y + 2} \right)^2} = 2$$

c) The radius of the circle is: $$R = \frac{{\left| {5.1 – 12.\left( { – 1} \right) – 1} \right|}}{{\sqrt {{5^) 2} + {{\left( { – 12} \right)}^2}} }} = \frac{{16}}{{13}}$$

The equation of the circle is: $${\left( {x – 1} \right)^2} + {\left( {y + 1} \right)^2} = {\left( {\frac{{16) }}{{13}}} \right)^2}$$

d) Let $$I\left( {a;b} \right)$$ be the midpoint of AB. So the coordinates of point I are: $$I\left( {1;1} \right)$$

The radius of the circle is: $$R = IA = \sqrt {{{\left( {3 – 1} \right)}^2} + {{\left( { – 4 – 1} \right)}^2 }} = \sqrt {29}$$

The equation of the circle is: $${\left( {x – 1} \right)^2} + {\left( {y – 1} \right)^2} = 29$$

e) Suppose the center of the circle is the point $$I\left( {a;b} \right)$$. We have: $$IA = IB = IC \Leftrightarrow I{A^2} = I{B^2} = I{C^2}$$

Since $$I{A^2} = I{B^2},I{B^2} = I{C^2}$$ then: $$\left\{ \begin{array}{l}{\ left( {1 – a} \right)^2} + {\left( {1 – b} \right)^2} = {\left( {3 – a} \right)^2} + {\left( {1 – b} \right)^2}\\{\left( {3 – a} \right)^2} + {\left( {1 – b} \right)^2} = {\left( { 0 – a} \right)^2} + {\left( {4 – b} \right)^2}\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a = 2\\b = 3\end{array} \right.$$ b

So $$I\left( {2;3} \right)$$ and $$R = IA = \sqrt {{{\left( { – 1} \right)}^2} + {{\left( { – 2} \right)}^2}} = \sqrt 5$$

So the equation of the circle passing through 3 points A,B,C is: $${\left( {x – 2} \right)^2} + {\left( {y – 3} \right)^2} = 5$$

### Solve the exercises Lesson 4, page 92, Math textbook 10 Kite episode 2

Make an equation of the tangent at the point with coordinate 3 on the circle $${\left( {x + 2} \right)^2} + {\left( {y + 7} \right)^2} = 169$$.

Solution method

Let the point ($${M_o}\left( {{x_o};{\rm{ }}{y_o}} \right)$$) lie on the circle (C) center I(a; b) radius R. Let $$\Delta$$ be the tangent at $${M_o}\left( {{x_o};{\rm{ }}{y_o}} \right)$$ on the circle. Then the equation of tangent $$\Delta$$ is:

$$\left( {{x_o} – a} \right)\left( {x – {x_o}} \right) + \left( {{y_o} – b} \right)\left( {y – {y_o }} \right) = 0$$

Solution guide

The coordinates of the point of contact are: $${M_1}\left( {3;5} \right),{M_2}\left( {3; – 12} \right)$$

The equation of the tangent to the circle passing through $${M_1}$$ is: $$– 5\left( {x – 3} \right) – 12\left( {y – 5} \right) = 0 \Leftrightarrow – 5x – 12y + 75 = 0$$

The equation of the tangent to the circle passing through $${M_2}$$ is:

$$– 5\left( {x – 3} \right) + 19(y + 12) = 0 \Leftrightarrow – 5x + 19y + 243 = 0$$

### Solution of Exercises Lesson 5, page 92, Math Textbook 10, Kite episode 2

Find m such that the line 3x + 4y + m = 0 touches the circle: $${\left( {x + 1} \right)^2} + {\left( {y-2} \right)^ 2} = 4$$.

Solution method

In the coordinate plane Oxy, for the line $$\Delta$$ there is the equation $${\rm{a}}x + by + c = 0\left( {{a^2} + {b^2} ) > 0} \right)$$ and the point $$M\left( {{x_o};{y_0}} \right)$$. The distance from the point M to the line $$\Delta$$, denoted by $$d\left( {M,\Delta } \right)$$ is calculated by the formula: $$d\left( {M, \Delta } \right) = \frac{{\left| {{\rm{a}}{x_o} + b{y_o} + c} \right|}}{{\sqrt {{a^2} + { b^2}} }}$$

Solution guide

For a line to touch the circle, $$d\left( {I,\Delta } \right) = R \Leftrightarrow \frac{{\left| {3.\left( { – 1} \right) + 4.2 + m} \right|}}{{\sqrt {{3^2} + {4^2}} }} = 2 \Leftrightarrow \left[\begin{array}{l}m=5\\m=–15\end{array}\right$$[\begin{array}{l}m=5\\m= –15\end{array}\right\)

### Solving exercises Lesson 6, page 92, Math textbook 10 Kite, episode 2

Figure 46 simulates a mobile phone base station located at position 1 with coordinates (- 2 ; 1) in the coordinate plane (unit on two axes is kilometers).

a) Make a circle equation describing the outer boundary of the coverage area, given that the base station is designed with a coverage radius of 3 km.

b) If the phone user is in the location with coordinates (-1; 3), can the service of this station be used? Explain.

c) Calculated according to the crow’s flight path, determine the shortest distance for a person at a location with coordinates (-3;4) to move to the coverage area in kilometers (round the result to the nearest point). tenths).

Solution method

The circle with center $$I\left( {a;b} \right)$$ and radius R has the equation: $${\left( {x – a} \right)^2} + {\left ( {y – b} \right)^2} = {R^2}$$

Solution guide

a) The equation of the circle describing the outer boundary of the coverage area is: $${\left( {x + 2} \right)^2} + {\left( {y – 1} \right)^2 } = 9$$

b) Distance from center I to A is: $$IA = \sqrt {{{\left( { – 1 + 2} \right)}^2} + {{\left( {3 – 1} \right) }^2}} = \sqrt 5$$

Since $$IA < 3$$, point A lies inside the boundary circle. So person A can serve the station.

c) Distance from center I to B is: $$IB = \sqrt {{{\left( { – 3 + 2} \right)}^2} + {{\left( {4 – 1} \right) }^2}} = \sqrt {10}$$

The shortest distance as the crow flies for a person in B to travel to the coverage area is:

$$IB – R = \sqrt {10} – 3\left( {km} \right)$$

### Solving exercises Exercise 7 page 92 Math textbook 10 Kite episode 2

Disc throw is a sport that competes in the Summer Olympic Games. When making a throw, the athlete usually turns his back to the direction of the throw, then rotates counterclockwise one and a half times in a circle to gain momentum and then releases his hand from the disc. Assume the disk moves on a circular path centered $$I\left( {0;\frac{3}{2}} \right)$$ of radius 0.8 in the coordinate plane Oxy (units on two axes). is meters). At the point$$M\left( {\frac{{\sqrt {39} }}{2};2} \right)$$, the disc is thrown away (Figure 47). In the first seconds immediately after being thrown, what is the equation of the disk’s trajectory?

Solution method

Let the point ($${M_o}\left( {{x_o};{\rm{ }}{y_o}} \right)$$) lie on the circle (C) center I(a; b) radius R. Let $$\Delta$$ be the tangent at $${M_o}\left( {{x_o};{\rm{ }}{y_o}} \right)$$ on the circle. Then the equation of tangent $$\Delta$$ is:

$$\left( {{x_o} – a} \right)\left( {x – {x_o}} \right) + \left( {{y_o} – b} \right)\left( {y – {y_o }} \right) = 0$$

Solution guide

After being thrown, the disk’s trajectory lies on the tangent to the circle with center I at point M.

So the disk’s orbital motion lies on a straight line whose equation is:

$$\begin{array}{l}\left( {\frac{{\sqrt {39} }}{{10}} – 0} \right)\left( {x – \frac{{\sqrt {39) } }}{{10}}} \right) + \left( {2 – \frac{3}{2}} \right)\left( {y – 2} \right) = 0\\ \Leftrightarrow \frac {{\sqrt {39} }}{{10}}\left( {x – \frac{{\sqrt {39} }}{{10}}} \right) + \frac{1}{2}\ left( {y – 2} \right) = 0\\ \Leftrightarrow \sqrt {39} x + 5y – 13.9 = 0\end{array}$$