Thursday, February 15, 2018

Chapter 34.8 - The Equation of Circle

In the previous section we saw the slopes of perpendicular lines. We also saw some examples. In this section we will see circles.

• We have seen that y = mx + c is the equation of a line. What does it mean?
Let us analyze:
1. Consider a line whose equation is y = mx + c
2. Take any point 'P' on the line 
• In the place of 'x' in the equation, put the x coordinate of the point  
• In the place of 'y' in the equation, put the y coordinate of the point  
3. Then the two sides of the equation will become equal. That is:
■ The coordinates of any point on the line will satisfy the equation of that line.
We can write the converse also:
■ If the coordinates of a point satisfy the equation of a line, then that point will lie on that line

Now consider a circle. Can we form a equation for that circle?
Let us try:
• There will be a large number of points in the circumference of a circle
    ♦ The coordinates of all those points should satisfy the equation of that circle
Let us see an example:
1. Consider the yellow circle in fig.34.29 below:
Fig.34.29
• It's center is at C(1,4). It's radius is 2 units
2. Consider any point P(x,y) on the circle. Then, using the distance formula, 
CP = [(x2-x1)2 + (y2-y1)]2 = [(x-1)2 + (y-4)2]
3. But this distance CP is the radius of the circle
So we can write:
[(x-1)2 + (y-4)2] = r = 2
4. Squaring both sides, we get:
[(x-1)2 + (y-4)2] = 4
5. The above equation can be taken as the equation of the circle because, every point on the circle will satisfy it. 
• This is because, it is simply the square of the distance between the 'center of the circle'  and 'the point on the circle'.
• And the distance from the center will be the same for which ever point we take.
An example
(i) The point with coordinates (1,2) lies on the circle. This is shown in the fig.
(ii) The distance between C and that point = [(x2-x1)2 + (y2-y1)2] 
[(1-1)2 + (2-4)2[(0)2 + (-2)2] = [4] = 2 units    
(iii) So 'square of the distance PO' = PO2 = 22= 4

The general case:
1. To write the general case, we consider a circle whose center is at any point C(x1,y1), and radius 'r' units
2. Then square of the distance between the center and any point (x,y) on the circle is:
[(x-x1)2 + (y-y1)2]
• But this distance is the square of the radius which is r2
3. Equating the two, we get:
[(x-x1)2 + (y-y1)2] = r2
• This is the general form of the equation of a circle
• Note the pattern:
    ♦ x1, which is the x coordinate of the center is subtracted from x
    ♦ y1, which is the y coordinate of the center is subtracted from y
4. From this we get the general form of the equation of any circle whose center (x1,y1) is at the origin O
• All we need to do is, put x1 = 0 and y1 = 0 in (3)
• Then we get: [x2 + y2] = r2
• This is shown in fig.34.30 below:
Fig.34.30
From this fig., we can see that:
• The distance from the center of the circle to  any point on the circle is indeed '[x2 + y2]' if the center is at the origin

Now we will see some solved examples:

Solved example 34.16
Find the equation of the line joining (1,2) and (2,4). In this, find the sequence of y coordinates of those points with the consecutive natural numbers 3, 4, 5, . . . as the x coordinates
Solution:
1. Equation of the line joining (1,2) and (2,4):
• Slope of the line =
m = (y2-y1)(x2-x1)  (4-2)(2-1) 21 = 2
• 'c' of the line = (y1-mx1) = [2 - (2 × 1)] = [2 - 2] = 0
• So equation of the line is
y = mx + c:
y = 2 × x + 0 ⟹ y = 2x
2. We know that the consecutive natural numbers form a sequence. That sequence is:
1, 2, 3, 4, 5, . . .
3. The terms of this sequence form the x coordinates of some points. We can write them as:
(1, _ ), (2, _ ), (3, _ ), (4, _ ), (5, _ ), . . .
4. The y coordinates in the above sequence are left blank. We have to find their values.
• It is given that those y coordinates form a sequence
5. It is easy to find the terms of that sequence. Because we have the equation of the line. It is: y = 2x
• From the equation it is clear that, what ever be the value of x, the value of y will be twice that x value. 
• So we can fill up the blank spaces in (3). We get:
(1,2), (2,4), (3,6), (4,8), (5,10), . . .
• Note that, the two points given to us in the question are the first two terms of the sequence.
• The line and the actual positions of the points are shown in the fig.34.31 below:
Fig.34.31
Solved example 34.17
Find the equation of the line joining (-1,3) and (2,5). Prove that if (u,v) is a point on this line, so is (u+3, v+2)
Solution:
1. Equation of the line joining (-1,3) and (2,5)
• Slope of the line =
m = (y2-y1)(x2-x1)  (5-3)(2-(-1)) 23
• 'c' of the line = (y1-mx1) = [3 - (23 × -1)] = [3 + 23] = 113
• So equation of the line is
y = mx + c:
y = 23 × x + 113 
2. Given that, (u,v) is a point on the line. What does that mean?
• It simply means that, any point on the line can be taken as (u,v)
• For example, if we consider the point (-1,3), we can take u = -1 and v = 3
• If we consider the point (2,5), we can take u = 2 and v = 5
3. Now we have prove that (u+3, v+2) is also a point on the line.
• For example, we know that (-1,3) is a point on the line. 
    ♦ Then (-1+3, 3+2) should also be a point on the line. 
    ♦ That is., (2,5) should also be a point on the line
• Similarly, we know that (2,5) is a point on the line. 
    ♦ Then (2+3, 5+2) should also be a point on the line. 
    ♦ That is., (5,7) should also be a point on the line 
• We have to prove it for the general case.
4. Consider the rough sketch in fig 34.32 below:
Fig.34.32
• A possible position of (u+3, v+2) is shown.
• The slope of the line joining (u,v) and (u+3, v+2) is
m = (y2-y1)(x2-x1) (v+2-v)(u+3-u) 23
• But this is same slope of the line through (-1,3), (2,5) and (u,v)
• Also, (u,v) is a common point. So (u+3, v+2) will lie in the same line

Solved example 34.18
Prove that for any number u, the point (u, 2u+3) is on the line joining (-1,1), (2,7).
Solution:
1. Consider the rough sketch in fig.34.33 below
Fig.34.33
• We can mark any point P' on the x axis. 
• If we draw a vertical line through that point it will surely meet the given line at some point. 
2. Let the marked point P' on the x axis have an x coordinate of 'u'
• Then the meeting point P will also have the x coordinate 'u'
3. The y coordinate of P is not known
■ We have to prove that the y coordinate at the meeting point is:
[(2 times the x coordinate) + 3]. 
• In other words, 
We have to prove that the y coordinate at the meeting point is:
2u+3
4. The equation of a line will help us to find the x coordinate when y coordinate is known
• The equation of a line will also help us to find the y coordinate when the x coordinate is known
5. So let us find the equation of the line:
Equation of the line joining (-1,1) and (2,7)
• Slope of the line =
m = (y2-y1)(x2-x1)  (7-1)(2-(-1)) 6= 2
• 'c' of the line = (y1-mx1) = [1 - (2 × -1)] = [1 + 2] = 3
• So equation of the line is
y = mx + c:
y = 2x + 3
6. From the equation it is clear that, at any point, the y coordinate will be:
[(2 times the x coordinate at that point) + 3] 
Hence proved

Solved example 34.19
The x coordinate of a point on the slanted line in fig.34.34(a) is 3
Fig.34.34
(i) What is it's y coordinate?
(ii) What is the slope of the line?
(iii) Write the equation of the line
Solution:
Part (i):
1. Drop a perpendicular from B on to the x axis. Let B' be the foot of the perpendicular. This is shown in fig.b
• So BB' will be the y coordinate of B
2. The ABB' is a 30o 60o triangle. We can find it's sides using trigonometric properties. (Details here)
• But for that, at least one side of the triangle should be known.
We have:
    ♦ Length OB' = 3 units
    ♦ Length OA = 1 unit (since x coordinate of A = 1)
    ♦ So AB' = 3 - 1 = 2 units
So now we have one side. We can apply the properties of a 30o 60o triangle 
• tan 60 = opposite sideadjacent side BB'AB'  BB'3
⟹ BB' = 23
Thus y coordinate of B = BB' = 23
Part (ii):
• Slope of the line =
m = (y2-y1)(x2-x1)  (2√3-0)(3-1) (2√3)(2) 3
Another method:
• If we know the angle θ that a line makes with the horizontal, then the slope of that line is simply tanθ.
• We discussed about it earlier in this chapter. Details here.
• For our present case, slope = tan 60 = 3
Part (iii):
Equation of the line joining (1,0) and (3,23)
• Slope of the line = m = 3
• 'c' of the line = (y1-mx1) = [0 - (3 × 1)] = [0 - 3] = -3
• So equation of the line is
y = mx + c:
y = 3x - ⟹ y = 3(x-1)

Solved example 34.20
In the fig.34.35(a), ABCD is a square.
Fig.34.35
Prove that for any point on the diagonal BD, the sum of x and y coordinates is zero
Solution:
1. The vertices A and C are diagonally opposite. If we are given two such points in a rectangle or a square, we can easily find the coordinates of the other two vertices. Details here.
• So the unknown coordinates at B and D are shown in fig.b
2. Now we have two points on the line BD. We can write the equation
• Equation of the line joining (2,-2) and (-2,2)
• Slope of the line =
m = (y2-y1)(x2-x1)  (2-(-2))(-2-2) 4-2 = -2
• 'c' of the line = (y1-mx1) = [-2 - (-2 × 2)] = [-2 + 4] = 2
• So equation of the line is
y = mx + c:

y = -2x + 2

Solved example 34.21
Prove that for any point on the line intersecting the axes in the fig.34.36 below, the sum of the x and y coordinates is 3
Fig.30.36
Solution:
1. We have the coordinates of two points on the line.
• Using them we can write the equation of the line
• Equation of the line joining (3,0) and (0,3)
• Slope of the line =
m = (y2-y1)(x2-x1)  (3-0)(0-3) 3-3 = -1
• 'c' of the line = (y1-mx1) = [0 - (-1 × 3)] = [0 - (-3)] = 3
• So equation of the line is
y = mx + c:
y = -1x + 3
2. From this equarion we get: x+y = 3
That is., for every point on this line, the sum of x and y coordinates will be 3

Solved example 34.22
Find the equation of the circle with center at the origin and radius 5 cm. Write the coordinates of 8 points on this circle
Solution:
1. We have seen at the beginning of this section that:
Equation of a circle with center O and radius r is
[x2 + y2] = r2
• So for our present case, the equation is: [x2 + y2] = 52  [x2 + y2] = 25
2. The circle cuts the axes at the following four points:
(i) P at the positive side of the x axis
(ii) P' at the negative side of the x axis
(iii) Q at the positive side of the y axis
(ii) Q' at the negative side of the y axis
These are shown in the fig.34.37(a) below:
We can find the coordinates of symmetric points in the Cartesian plane with out any calculations.
Fig.34.37
• Now we will calculate the coordinates of each of the above four points
(i) Point P: Since the point P is on the x axis, the y coordinate will be equal to zero. 
• So put y = 0 in the equation of the circle in (1)
• We get: [x2 + 0] = 25  ⟹ x2 = 25 ⟹ x = 25 = ±5
• There will be two roots for 25. They are +5 and -5.
• That means, the circle will intersect with the x axis at two points. 
    ♦ One point is P on the positive side of the x axis. The coordinates are (5,0)
    ♦ The other point P' is on the negative side of the x axis. The coordinates are (-5,0)
• This is shown in the fig.30.37(a) above
• So we started out to find P and got both P and P'
(ii) Point P': This is already obtained above
(iii) Point Q: Since the point Q is on the y axis, the x coordinate will be equal to zero. 
• So put x = 0 in the equation of the circle in (1)
• We get: [0 + y2] = 25  ⟹ y2 = 25 ⟹ y = 25 = ±5
• There will be two roots for 25. They are +5 and -5.
• That means, the circle will intersect with the y axis at two points. 
    ♦ One point Q is on the positive side of the y axis. The coordinates are (0,5)
    ♦ The other point Q' is on the negative side of the y axis. The coordinates are (0,-5)
• This is shown in the fig.30.37(a) above
• So we started out to find Q and got both Q and Q'
(ii) Point Q': This is already obtained above
3. We want four more points. 
• For that, we draw a line at an angle of 60o with the positive side of the x axis. 
    ♦ This line meets the circle at R. This is shown in fig.34.37(b) above
• A perpendicular is dropped from R on to the x axis. The foot of the perpendicular is R'
    ♦ Now x coordinate of R will be equal to OR'
    ♦ y coordinate of R will be equal to RR'
• This is a problem similar to solved example 34.19 that we saw above. In that problem we used the trigonometrical ratio tan. In this example we will use sine and cosine
Consider ORR'. We have:
• sin 60 = opposite sidehypotenuse RR'OR  RR'
But from the tables, sin 60 = √32
So we can write: RR'√32 ⟹ RR' = (5√3)2
• cos 60 = adjacent side hypotenuse OR'OR  OR'
But from the tables, cos 60 = 12
So we can write: OR'12 ⟹ OR' = 52 
Thus the coordinates of R are: [5(5√3)2]
Check:
• Applying Pythagoras theorem in ORR', we have:
OR [(OR')2 + (RR')2[(52)2 + ((5√3)2)2] = [254 + 754] = [1004] = [25] = 5 units
• The radius is given as 5 cm. So our calculations are correct
4. Three more points can be obtained in the same way. They are: S, S'' and R''. They are shown in fig.b. They are symmetrical to the point R that we calculated above. So the coordinates can be written easily:
Point S: [-5(5√3)2]
Point S'': [-5(-5√3)2]
Point R'': [5(-5√3)2]


Solved example 34.23 
Let (x,y) be a point on the circle with the line joining (0,1) and (2,3) as diameter. 
(i) Prove that x2+ y2 - 2x - 4y +3 = 0
(ii) Find the coordinates of the points where the circle cuts the x axis
Solution:
1. The end points of a diameter are (0,1) and (2,3)
From this we get: 
• x coordinate of the midpoint = (x1+x2)2 = (0+2)2= 1
• y coordinate of the midpoint = (y1+y2)2 = (1+3)4= 2
• So the coordinates of the midpoint are (Details here): (1,2)
• But the midpoint of the diameter is same as the center of the circle
2. Length of the diameter can be calculated using the distance formula:
Diameter = [(x2-x1)2 + (y2-y1)]2 = [(2-0)2 + (3-1)2[(2)2 + (2)2] = 8 = [2×4] = 22
• So radius = (2√2)2 = 2 units
3. The general form of the equation of any circle whose center is at any point C(x1,y1), and radius is 'r' units is: [(x-x1)2 + (y-y1)2] = r2.
• So we get the equation of our circle:
[(x-1)2 + (y-2)2] = (2)2.
• Expanding the above expression, we get:
x2 -2x + 1 + y2 - 4y + 4 = 2 ⟹ x2+ y2 - 2x - 4y +3 = 0
Part (ii):
1. At the points where the circle cuts the x axis, the y coordinates will be zero.
2. So we can put y = 0 in the equation of the circle:
x2+ 02 - 2x - 4×+3 = 0 ⟹ x2 - 2x +3 = 0
3. This is of the form ax2 + bx + c = 0 (Details here)
Where: a = 1, b = (-2) and c = 3
4. So we can use the general formula to solve the equation
5. b2-4ac = (-2)2-4×1×3 = 4 - 12 = -8
• This is a negative quantity.
• So '√[b2-4ac]' cannot be calculated
• The equation does not have a solution. That means, the circle never crosses the x axis.
• The actual positions in the Cartesian plane are shown in the fig.34.38 below:
Fig.34.38
• Note that the radius is 2 units. It is difficult to mark the value of 2 precisely on the x or y axis
• But we have seen the methods to draw a line of length 2 geometrically (Details here)
• Using a graph paper, it is even more easier because, 2 is the diagonal of a square of side 1 unit

Solved example 34.24
What is the equation of the circle in the fig.34.39 below?
Fig.34.39
Solution:
• We know that, the axes are always perpendicular to each other
• In this problem, the point of intersection of the axes, that is., the origin, lies on the circle
• So AB will be a diameter of the circle (Details here)
1. The end points of a diameter are (0,2) and (4,0)
From this we get: 
• x coordinate of the midpoint = (x1+x2)2 = (0+4)4= 2
• y coordinate of the midpoint = (y1+y2)2 = (2+0)2= 1
• So the coordinates of the midpoint are (Details here): (2,1)
• But the midpoint of the diameter is same as the center of the circle
2. Length of the diameter can be calculated using the distance formula:
Diameter = [(x2-x1)2 + (y2-y1)]2 = [(4-0)2 + (0-2)2[(4)2 + (-2)2] = √20 = [5×4] = 2√5
• So radius = (2√5)2  5 units
3. The general form of the equation of any circle whose center is at any point C(x1,y1), and radius is 'r' units is: [(x-x1)2 + (y-y1)2] = r2.
• So we get the equation of our circle:
[(x-2)2 + (y-1)2] = (5)2.
• Expanding the above expression, we get:
x2 -4x + 4 + y2 - 2y + 1 = 5 ⟹ x2+ y2 - 4x - 2y = 0


In the next section, we will see Polynomials.


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