In the previous section we saw some solved examples using the Distance formula. In this section we will see a few more solved examples.
Solved example 31.29
Lines are drawn from a point inside a rectangle to all the four vertices. The lengths of three of those lines are shown in fig.31.51(a) below. Find the length of the fourth line.
Solution:
1. Given that it is a rectangle. So it's sides will be parallel to the coordinate axes.
2. We will place it in the coordinate plane in such a way that, the left bottom corner coincides with the origin. Let the name of the rectangle be OABC. This is shown in fig.31.51(b).
• The length OA coincides with the x-axis
• The width OC coincides with the y-axis
3. Coordinates of A:
• Coordinates of any point on the x-axis will have a general form: (x,0). Here, 'x' can take any value.
• Our rectangle has a definite length. For point A, 'x' will be a constant. So we will put 'a' in the place of 'x'. So the point A has coordinates: (a,0).
4. Coordinates of C:
• Coordinates of any point on the y-axis will have a general form: (0,y). Here, 'y' can take any value.
• Our rectangle has a definite width. For point C, 'y' will be a constant. So we will put 'c' in the place of 'y'. So the point C has coordinates: (0,c).
5. Coordinates of B:
• From the coordinates of A and C, we can easily write:
The coordinates of B are: (a,c)
6. Let P be the point inside the rectangle. We have to find the distance PB
7. Let the coordinates of P be (x,y)
• Then we get:
PB = √[(x2-x1)2 + (y2-y1)2] = √[(a-x)2 + (b-y)2]
8. We will now write the known distances:
(i) OP is the distance of P(x,y) from the origin. So OP = 4 = √[x2 + y2]
(ii) AP = 5 cm = √[(a-x)2 + (0-y)2] = √[(a-x)2 + y2]
(iii) CP = 3 cm = √[(0-x)2 + (b-y)2] = √[x2 + (b-y)2]
9. Let us square the distances:
(i) OP2 = [x2 + y2] = 16
(ii) AP2 = [(a-x)2 + y2] = 25
(iii) CP2 = [x2 + (b-y)2] = 9
(iv) PB2 = [(a-x)2 + (b-y)2]
10. Adding (ii) and (iii) we get: AP2 + CP2 = [(a-x)2 + y2] + [x2 + (b-y)2] = 25 + 9 = 34
• Rearranging the above equation, we get:
[(a-x)2 + (b-y)2 + x2 + y2] = 34
11. But from 9(i), x2 + y2 = 16
• Substituting this in (10) we get: [(a-x)2 + (b-y)2 + 16] = 34
⟹ [(a-x)2 + (b-y)2] = 18
• Taking square roots we get: √[(a-x)2 + (b-y)2] = √18
12. But from (7), what we want is: '√[(a-x)2 + (b-y)2]'
So we can write:
PB = √[(a-x)2 + (b-y)2] = √18 cm
Solved example 31.30
Calculate the lengths of the four sides and the two diagonals of the quadrilateral below:
Solution:
1. Distance AB:
• We can use the distance formula directly.
• But AB is a line parallel to y-axis (∵ the x coordinates of A and B are the same)
• So it is better to use the simpler method:
AB = |(y2-y1)| = |(-2-1)| = |(-3)| = 3 units
2. Distance BC:
• We can use the distance formula directly.
• But BC is a line parallel to x-axis (∵ the y coordinates of B and C are the same)
• So it is better to use the simpler method:
BC = |(x2-x1)| = |(1-(-3))| = |(4)| = 4 units
3. Distance OC:
• We can use the distance formula directly.
• But OC is the distance of C from the origin O
• So it is better to use the simpler method:
OC = √[x2 + y2] = √[12 + (-2)2] = √[1 + 4] = √5
4. Distance OA:
• We can use the distance formula directly.
• But OA is the distance of A from the origin O
• So it is better to use the simpler method:
OA = √[x2 + y2] = √[12 + (-3)2] = √[1 + 9] = √10
Solved example 31.31
Prove that, by joining points (2,1), (3,4), (-3,6) we get a right triangle
Solution:
1. Let the points be A, B and C. There are three possible distances. Let us find all of them:
• AB = √[(x2-x1)2 + (y2-y1)2] = √[(3-2)2 + (4-1)2] = √[1 + 9] = √[10]
• BC = √[(-3-3)2 + (6-4)2] = √[36 + 4] = √[40]
• AC = √[(-3-2)2 + (6-1)2] = √[25 + 25] = √[50]
2. Let us square the distances:
• AB2 = 10 • BC2 = 40 • AC2 = 50
3. We get: AB2 + BC2 = AC2.
Based on the Pythagoras theorem, AC is the hypotenuse and AB and BC are the legs. So the triangle got by joining the given points is a right triangle.
Solved example 31.32
A circle of radius 10 units is drawn with the origin as center.
(i) Check whether each of the points with coordinates (6,9), (5,9), (6,8) is inside, outside or on the circle
(ii) Write the coordinates of 8 points on this circle
Solution:
Part (i):
1. Distance between origin and (6,9)
= √[x2 + y2] = √[62 + 92] = √[36 + 81] = √117
• √117 is greater than 10. [∵ (√117)2 = 117 AND 102 = 100]
• So (6,9) lies out side the circle
2. Distance between origin and (5,9)
= √[x2 + y2] = √[52 + 92] = √[25 + 81] = √106
• √106 is greater than 10. [∵ (√106)2 = 106 AND 102 = 100]
• So (5,9) lies out side the circle
3. Distance between origin and (6,8)
= √[x2 + y2] = √[62 + 82] = √[36 + 64] = √100 = 10
• So (6,8) lies on the circle
The three points are shown in the fig.31.53 below:
Part (ii):
1. We are asked to find 8 points. 4 points can be found out very easily:
(i) Consider (fig.31.53 above) the point at which the circle intersects the positive side of the x-axis
(ii) This point is 'a point on the circle'. But any point on the circle will be at a distance of 10 units from O
(iii) So this 'point of intersection' is at a distance of 10 units from O
(iv) Since this point is on the x-axis, it's coordinates will be (10,0)
(v) Using the same steps, we can find three more points where the circle intersects with the axes:
• (-10,0) in the negative side of the x-axis
• (0,10) in the positive side of the y-axis
• (0,-10) in the negative side of the y-axis
2. Now we want 4 more points. Consider fig.31.54 below:
(i) Draw a line at an angle of 30o with the positive side of the x-axis
• Let it intersect the circle at A
(ii) Drop a perpendicular from A
• Let the foot of this perpendicular on the x-axis be B
(iii) Then OAB is a 30o, 60o triangle. We have seen the details of such triangles here.
(iv) The hypotenuse will be 2 times the smallest side
• The smallest side is AB (∵ AB is opposite the smallest angle 30o)
• So OA = 2×AB ⟹ AB = OA⁄2 = 10⁄2 = 5 units
• So distance of A from the x-axis is 5
• So y coordinate of A is 5
(v) In the right triangle OAB, the altitude will be √3 times the smallest side
• So OB = √3×AB = 5√3
• So distance of A from the y-axis is 5√3
• So x coordinate of A is 5√3
(vi) From (iv) and (v) we get: Coordinates of A are: (5√3,5)
(vii) Using the same steps, we can find 3 more points in the second, third and fourth quadrants:
• (-5√3,5) in the second quadrant
• (-5√3,-5) in the third quadrant
• (5√3,-5) in the fourth quadrant
■ So the required 8 points are: (10,0), (-10,0), (0,10), (0,-10), (5√3,5), (-5√3,5), (-5√3,-5) and (5√3,-5)
Solved example 31.33
Find the coordinates of the point where a circle of radius √2, centered on the point with coordinates (1,1) cut the axes.
Solution:
1. Consider the circle in fig.31.55(a) below. It's center is at 'C' which has coordinates (1,1)
• A line is drawn from the origin O to C. Also a perpendicular CA is dropped from C onto the x-axis
2. Since the coordinates of C are (1,1), we get:
• OA = AC = 1 unit
3. ⊿OAC is a right triangle. Applying Pythagoras theorem, we get:
OC = √[OA2 + AC2] = √[12 + 12] = √[2]
4. But it is given that, the radius of the circle is √2
• So 'O' is one of the points where the circle cut the axes.
5. We have to find the other two points B and D also:
• Draw BC as shown in fig.31.55(b). Now we have one more right triangle: ⊿ABC
• It's hypotenuse is BC. But BC is a radius. So we get: BC = √2
6. Applying Pythagoras theorem, we get:
• AB = √[BC2 - AC2] = √[(√2)2 -12] = √[2-1] = √1 = 1
• So OB = OA + AB = 1 + 1 = 2 units
• Thus the coordinates of B are: (2,0)
7. Next we want the coordinates of D at which the circle cuts the y-axis:
• Draw a horizontal CE through C. Also draw DC. This is shown in fig.31.55(c)
• Now we have two right triangles: ⊿OCE and ⊿DCE
8. Consider ⊿DCE
• Since the coordinates of C are (1,1), we get:
• OE = CE = 1 unit.
• CD = √2 (∵ CD is a radius)
9. Applying Pythagoras theorem, we get:
• ED = √[CD2 - CE2] = √[(√2)2 -12] = √[2-1] = √1 = 1
• So OD = OE + ED = 1 + 1 = 2 units
• Thus the coordinates of D are: (0,2)
10. So the circle cuts the axes at: (0,0), (2,0) and (0,2)
Solved example 31.34
The coordinates of the vertices of a triangle are (1,2), (2,3) and (3,1). Find the coordinates of the center of it's circumcircle and the circumradius
Solution:
1. The circumcircle of a triangle is the circle which passes through all the three vertices of a triangle. (Details here)
• The center of that circle is called the circumcenter.
• So this circumcenter will be equidistant from any point on the circle.
• So the three vertices will be equidistant from the circumcenter.
2. Let the circumcenter 'C' have the coordinates: (x,y)
• Let the vertices be P, Q and R
3. Then distant from the first point = CP = √[(x-1)2 + (y-2)2]
• Squaring both sides we get:
CP2 = [(x-1)2 + (y-2)2] ⟹ CP2 = x2 - 2x + 1 + y2 - 4y + 4 = x2 + y2 - 2x - 4y + 5
4. Distant from the second point = CQ = √[(x-2)2 + (y-3)2]
• Squaring both sides we get:
CQ2 = [(x-2)2 + (y-3)2] ⟹ CQ2 = x2 - 4x + 4 + y2 - 6y + 9 = x2 + y2 - 4x - 6y + 13
5. Distant from the third point = CR = √[(x-3)2 + (y-1)2]
• Squaring both sides we get:
CR2 = [(x-3)2 + (y-1)2] ⟹ CR2 = x2 - 6x + 9 + y2 - 2y + 1 = x2 + y2 - 6x - 2y + 10
6. We know that CP = CQ = CR
• Squaring we get: CP2 = CQ2 = CR2
7. So equating (3) and (4) we get:
CP2 = CQ2 ⟹ x2 + y2 - 2x - 4y + 5 = x2 + y2 - 4x - 6y + 13
• Simplifying we get:
- 2x - 4y + 5 = - 4x - 6y + 13 ⟹ 2x + 2y = 8 ⟹ x + y = 4
8. Equating (4) and (5) we get:
CQ2 = CR2 ⟹ x2 + y2 - 4x - 6y + 13 = x2 + y2 - 6x - 2y + 10
• Simplifying we get:
- 4x - 6y + 13 = - 6x - 2y + 10 ⟹ 2x - 4y = -3
9. So we have two equations:
(i) From (7): x + y = 4
(ii) Fom (8): 2x - 4y = -3
10. From 9(i) we get: x = (4-y)
• Substituting this in 9(ii) we get: 2(4-y) - 4y = -3
⟹ 8 - 2y - 4y = -3 ⟹ 8-6y = -3 ⟹ 6y = 11 ⟹ y = 11⁄6
11. Substituting this value of y in 9(i) we get: x + 11⁄6 = 4
⟹ x = 4 - 11⁄6 = (24-11)⁄6 = 13⁄6
12. So the coordinates of the circumcenter are: (13⁄6 ,11⁄6)
13. Now we can find the circumradius. It is the distance between C and any one of the points P, Q or R
14. We will calculate CP:
Substituting for x and y in (3) we get:
CP2 = [(x-1)2 + (y-2)2] = [(13⁄6 -1)2 + (11⁄6 -2)2] = [((13-6)⁄6)2 + ((11-12)⁄6)2]
= [(7⁄6)2 + (-1⁄6)2] = [49⁄36 + 1⁄36] = 50⁄36
⟹ CP = √[50⁄36] = [(√25×√2)⁄(√36)] = [(5√2)⁄6]
15. The triangle and the circumcircle is shown in the fig.31.56 below:
Solved example 31.34
In the fig.31.57 below, the center of the circle is at the origin. A and B are two points on the circle. Calculate the length of the chord AB.
Solution:
1. In fig.31.57(b), a perpendicular is dropped from A
• Let the foot of this perpendicular on the x-axis be C
2. Then OAC is a 30o, 60o triangle. We have seen the details of such triangles here.
(i) The hypotenuse will be 2 times the smallest side
• The smallest side is AC (∵ AC is opposite the smallest angle 30o)
• So OA = 2×AC ⟹ AC = OA⁄2 = 2⁄2 = 1 unit
• So distance of A from the x-axis is 1
• So y coordinate of A is 1
(ii) In the right triangle OAC, the altitude will be √3 times the smallest side
• So OC = √3×AC = √3 units
• So distance of A from the y-axis is √3
• So x coordinate of A is √3
(iii) From (i) and (ii) we get: Coordinates of A are: (√3,1)
3. Drop a perpendicular from B
• Let the foot of this perpendicular on the x-axis be D
• We want ∠DOB. It can be calculated as follows:
∠DOB + ∠BOA + ∠AOC = 180o
⟹ ∠DOB + 90 + 30 = 180o ⟹ ∠DOB + 120 = 180o ⟹ ∠DOB = 60o
4. Then OAC is a 30o, 60o triangle.
(i) The hypotenuse will be 2 times the smallest side
• The smallest side is OD (∵ OD is opposite the smallest angle 30o)
• So OB = 2×OD ⟹ OD = OB⁄2 = 2⁄2 = 1 unit
• So distance of B from the y-axis is 1
• So x coordinate of B is -1 (∵ B is in the second quadrant)
(ii) In the right triangle OBD, the altitude will be √3 times the smallest side
• So BD = √3×OD = √3 units
• So distance of B from the x-axis is √3
• So y coordinate of B is √3
(iii) From (i) and (ii) we get: Coordinates of B are: (-1,√3)
5. Now we can calculate the length of the chord AB
AB = √[(-1-√3)2 + (√3-1)2] = √[(√3+1)2 + (√3-1)2] = √[3 + 2√3 +1 + 3 - 2√3 + 1]
= √[8] = √[4×2] = √4 × √2 = 2√2 units
Solved example 31.29
Lines are drawn from a point inside a rectangle to all the four vertices. The lengths of three of those lines are shown in fig.31.51(a) below. Find the length of the fourth line.
 |
| Fig.31.51 |
1. Given that it is a rectangle. So it's sides will be parallel to the coordinate axes.
2. We will place it in the coordinate plane in such a way that, the left bottom corner coincides with the origin. Let the name of the rectangle be OABC. This is shown in fig.31.51(b).
• The length OA coincides with the x-axis
• The width OC coincides with the y-axis
3. Coordinates of A:
• Coordinates of any point on the x-axis will have a general form: (x,0). Here, 'x' can take any value.
• Our rectangle has a definite length. For point A, 'x' will be a constant. So we will put 'a' in the place of 'x'. So the point A has coordinates: (a,0).
4. Coordinates of C:
• Coordinates of any point on the y-axis will have a general form: (0,y). Here, 'y' can take any value.
• Our rectangle has a definite width. For point C, 'y' will be a constant. So we will put 'c' in the place of 'y'. So the point C has coordinates: (0,c).
5. Coordinates of B:
• From the coordinates of A and C, we can easily write:
The coordinates of B are: (a,c)
6. Let P be the point inside the rectangle. We have to find the distance PB
7. Let the coordinates of P be (x,y)
• Then we get:
PB = √[(x2-x1)2 + (y2-y1)2] = √[(a-x)2 + (b-y)2]
8. We will now write the known distances:
(i) OP is the distance of P(x,y) from the origin. So OP = 4 = √[x2 + y2]
(ii) AP = 5 cm = √[(a-x)2 + (0-y)2] = √[(a-x)2 + y2]
(iii) CP = 3 cm = √[(0-x)2 + (b-y)2] = √[x2 + (b-y)2]
9. Let us square the distances:
(i) OP2 = [x2 + y2] = 16
(ii) AP2 = [(a-x)2 + y2] = 25
(iii) CP2 = [x2 + (b-y)2] = 9
(iv) PB2 = [(a-x)2 + (b-y)2]
10. Adding (ii) and (iii) we get: AP2 + CP2 = [(a-x)2 + y2] + [x2 + (b-y)2] = 25 + 9 = 34
• Rearranging the above equation, we get:
[(a-x)2 + (b-y)2 + x2 + y2] = 34
11. But from 9(i), x2 + y2 = 16
• Substituting this in (10) we get: [(a-x)2 + (b-y)2 + 16] = 34
⟹ [(a-x)2 + (b-y)2] = 18
• Taking square roots we get: √[(a-x)2 + (b-y)2] = √18
12. But from (7), what we want is: '√[(a-x)2 + (b-y)2]'
So we can write:
PB = √[(a-x)2 + (b-y)2] = √18 cm
Solved example 31.30
Calculate the lengths of the four sides and the two diagonals of the quadrilateral below:
 |
| Fig.31.52 |
1. Distance AB:
• We can use the distance formula directly.
• But AB is a line parallel to y-axis (∵ the x coordinates of A and B are the same)
• So it is better to use the simpler method:
AB = |(y2-y1)| = |(-2-1)| = |(-3)| = 3 units
2. Distance BC:
• We can use the distance formula directly.
• But BC is a line parallel to x-axis (∵ the y coordinates of B and C are the same)
• So it is better to use the simpler method:
BC = |(x2-x1)| = |(1-(-3))| = |(4)| = 4 units
3. Distance OC:
• We can use the distance formula directly.
• But OC is the distance of C from the origin O
• So it is better to use the simpler method:
OC = √[x2 + y2] = √[12 + (-2)2] = √[1 + 4] = √5
4. Distance OA:
• We can use the distance formula directly.
• But OA is the distance of A from the origin O
• So it is better to use the simpler method:
OA = √[x2 + y2] = √[12 + (-3)2] = √[1 + 9] = √10
Solved example 31.31
Prove that, by joining points (2,1), (3,4), (-3,6) we get a right triangle
Solution:
1. Let the points be A, B and C. There are three possible distances. Let us find all of them:
• AB = √[(x2-x1)2 + (y2-y1)2] = √[(3-2)2 + (4-1)2] = √[1 + 9] = √[10]
• BC = √[(-3-3)2 + (6-4)2] = √[36 + 4] = √[40]
• AC = √[(-3-2)2 + (6-1)2] = √[25 + 25] = √[50]
2. Let us square the distances:
• AB2 = 10 • BC2 = 40 • AC2 = 50
3. We get: AB2 + BC2 = AC2.
Based on the Pythagoras theorem, AC is the hypotenuse and AB and BC are the legs. So the triangle got by joining the given points is a right triangle.
Solved example 31.32
A circle of radius 10 units is drawn with the origin as center.
(i) Check whether each of the points with coordinates (6,9), (5,9), (6,8) is inside, outside or on the circle
(ii) Write the coordinates of 8 points on this circle
Solution:
Part (i):
1. Distance between origin and (6,9)
= √[x2 + y2] = √[62 + 92] = √[36 + 81] = √117
• √117 is greater than 10. [∵ (√117)2 = 117 AND 102 = 100]
• So (6,9) lies out side the circle
2. Distance between origin and (5,9)
= √[x2 + y2] = √[52 + 92] = √[25 + 81] = √106
• √106 is greater than 10. [∵ (√106)2 = 106 AND 102 = 100]
• So (5,9) lies out side the circle
3. Distance between origin and (6,8)
= √[x2 + y2] = √[62 + 82] = √[36 + 64] = √100 = 10
• So (6,8) lies on the circle
The three points are shown in the fig.31.53 below:
 |
| Fig.31.53 |
1. We are asked to find 8 points. 4 points can be found out very easily:
(i) Consider (fig.31.53 above) the point at which the circle intersects the positive side of the x-axis
(ii) This point is 'a point on the circle'. But any point on the circle will be at a distance of 10 units from O
(iii) So this 'point of intersection' is at a distance of 10 units from O
(iv) Since this point is on the x-axis, it's coordinates will be (10,0)
(v) Using the same steps, we can find three more points where the circle intersects with the axes:
• (-10,0) in the negative side of the x-axis
• (0,10) in the positive side of the y-axis
• (0,-10) in the negative side of the y-axis
2. Now we want 4 more points. Consider fig.31.54 below:
 |
| Fig.31.54 |
• Let it intersect the circle at A
(ii) Drop a perpendicular from A
• Let the foot of this perpendicular on the x-axis be B
(iii) Then OAB is a 30o, 60o triangle. We have seen the details of such triangles here.
(iv) The hypotenuse will be 2 times the smallest side
• The smallest side is AB (∵ AB is opposite the smallest angle 30o)
• So OA = 2×AB ⟹ AB = OA⁄2 = 10⁄2 = 5 units
• So distance of A from the x-axis is 5
• So y coordinate of A is 5
(v) In the right triangle OAB, the altitude will be √3 times the smallest side
• So OB = √3×AB = 5√3
• So distance of A from the y-axis is 5√3
• So x coordinate of A is 5√3
(vi) From (iv) and (v) we get: Coordinates of A are: (5√3,5)
(vii) Using the same steps, we can find 3 more points in the second, third and fourth quadrants:
• (-5√3,5) in the second quadrant
• (-5√3,-5) in the third quadrant
• (5√3,-5) in the fourth quadrant
■ So the required 8 points are: (10,0), (-10,0), (0,10), (0,-10), (5√3,5), (-5√3,5), (-5√3,-5) and (5√3,-5)
Solved example 31.33
Find the coordinates of the point where a circle of radius √2, centered on the point with coordinates (1,1) cut the axes.
Solution:
1. Consider the circle in fig.31.55(a) below. It's center is at 'C' which has coordinates (1,1)
 |
| Fig.31.55 |
2. Since the coordinates of C are (1,1), we get:
• OA = AC = 1 unit
3. ⊿OAC is a right triangle. Applying Pythagoras theorem, we get:
OC = √[OA2 + AC2] = √[12 + 12] = √[2]
4. But it is given that, the radius of the circle is √2
• So 'O' is one of the points where the circle cut the axes.
5. We have to find the other two points B and D also:
• Draw BC as shown in fig.31.55(b). Now we have one more right triangle: ⊿ABC
• It's hypotenuse is BC. But BC is a radius. So we get: BC = √2
6. Applying Pythagoras theorem, we get:
• AB = √[BC2 - AC2] = √[(√2)2 -12] = √[2-1] = √1 = 1
• So OB = OA + AB = 1 + 1 = 2 units
• Thus the coordinates of B are: (2,0)
7. Next we want the coordinates of D at which the circle cuts the y-axis:
• Draw a horizontal CE through C. Also draw DC. This is shown in fig.31.55(c)
• Now we have two right triangles: ⊿OCE and ⊿DCE
8. Consider ⊿DCE
• Since the coordinates of C are (1,1), we get:
• OE = CE = 1 unit.
• CD = √2 (∵ CD is a radius)
9. Applying Pythagoras theorem, we get:
• ED = √[CD2 - CE2] = √[(√2)2 -12] = √[2-1] = √1 = 1
• So OD = OE + ED = 1 + 1 = 2 units
• Thus the coordinates of D are: (0,2)
10. So the circle cuts the axes at: (0,0), (2,0) and (0,2)
Solved example 31.34
The coordinates of the vertices of a triangle are (1,2), (2,3) and (3,1). Find the coordinates of the center of it's circumcircle and the circumradius
Solution:
1. The circumcircle of a triangle is the circle which passes through all the three vertices of a triangle. (Details here)
• The center of that circle is called the circumcenter.
• So this circumcenter will be equidistant from any point on the circle.
• So the three vertices will be equidistant from the circumcenter.
2. Let the circumcenter 'C' have the coordinates: (x,y)
• Let the vertices be P, Q and R
3. Then distant from the first point = CP = √[(x-1)2 + (y-2)2]
• Squaring both sides we get:
CP2 = [(x-1)2 + (y-2)2] ⟹ CP2 = x2 - 2x + 1 + y2 - 4y + 4 = x2 + y2 - 2x - 4y + 5
4. Distant from the second point = CQ = √[(x-2)2 + (y-3)2]
• Squaring both sides we get:
CQ2 = [(x-2)2 + (y-3)2] ⟹ CQ2 = x2 - 4x + 4 + y2 - 6y + 9 = x2 + y2 - 4x - 6y + 13
5. Distant from the third point = CR = √[(x-3)2 + (y-1)2]
• Squaring both sides we get:
CR2 = [(x-3)2 + (y-1)2] ⟹ CR2 = x2 - 6x + 9 + y2 - 2y + 1 = x2 + y2 - 6x - 2y + 10
6. We know that CP = CQ = CR
• Squaring we get: CP2 = CQ2 = CR2
7. So equating (3) and (4) we get:
CP2 = CQ2 ⟹ x2 + y2 - 2x - 4y + 5 = x2 + y2 - 4x - 6y + 13
• Simplifying we get:
- 2x - 4y + 5 = - 4x - 6y + 13 ⟹ 2x + 2y = 8 ⟹ x + y = 4
8. Equating (4) and (5) we get:
CQ2 = CR2 ⟹ x2 + y2 - 4x - 6y + 13 = x2 + y2 - 6x - 2y + 10
• Simplifying we get:
- 4x - 6y + 13 = - 6x - 2y + 10 ⟹ 2x - 4y = -3
9. So we have two equations:
(i) From (7): x + y = 4
(ii) Fom (8): 2x - 4y = -3
10. From 9(i) we get: x = (4-y)
• Substituting this in 9(ii) we get: 2(4-y) - 4y = -3
⟹ 8 - 2y - 4y = -3 ⟹ 8-6y = -3 ⟹ 6y = 11 ⟹ y = 11⁄6
11. Substituting this value of y in 9(i) we get: x + 11⁄6 = 4
⟹ x = 4 - 11⁄6 = (24-11)⁄6 = 13⁄6
12. So the coordinates of the circumcenter are: (13⁄6 ,11⁄6)
13. Now we can find the circumradius. It is the distance between C and any one of the points P, Q or R
14. We will calculate CP:
Substituting for x and y in (3) we get:
CP2 = [(x-1)2 + (y-2)2] = [(13⁄6 -1)2 + (11⁄6 -2)2] = [((13-6)⁄6)2 + ((11-12)⁄6)2]
= [(7⁄6)2 + (-1⁄6)2] = [49⁄36 + 1⁄36] = 50⁄36
⟹ CP = √[50⁄36] = [(√25×√2)⁄(√36)] = [(5√2)⁄6]
15. The triangle and the circumcircle is shown in the fig.31.56 below:
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| Fig.31.56 |
In the fig.31.57 below, the center of the circle is at the origin. A and B are two points on the circle. Calculate the length of the chord AB.
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| Fig.31.57 |
1. In fig.31.57(b), a perpendicular is dropped from A
• Let the foot of this perpendicular on the x-axis be C
2. Then OAC is a 30o, 60o triangle. We have seen the details of such triangles here.
(i) The hypotenuse will be 2 times the smallest side
• The smallest side is AC (∵ AC is opposite the smallest angle 30o)
• So OA = 2×AC ⟹ AC = OA⁄2 = 2⁄2 = 1 unit
• So distance of A from the x-axis is 1
• So y coordinate of A is 1
(ii) In the right triangle OAC, the altitude will be √3 times the smallest side
• So OC = √3×AC = √3 units
• So distance of A from the y-axis is √3
• So x coordinate of A is √3
(iii) From (i) and (ii) we get: Coordinates of A are: (√3,1)
3. Drop a perpendicular from B
• Let the foot of this perpendicular on the x-axis be D
• We want ∠DOB. It can be calculated as follows:
∠DOB + ∠BOA + ∠AOC = 180o
⟹ ∠DOB + 90 + 30 = 180o ⟹ ∠DOB + 120 = 180o ⟹ ∠DOB = 60o
4. Then OAC is a 30o, 60o triangle.
(i) The hypotenuse will be 2 times the smallest side
• The smallest side is OD (∵ OD is opposite the smallest angle 30o)
• So OB = 2×OD ⟹ OD = OB⁄2 = 2⁄2 = 1 unit
• So distance of B from the y-axis is 1
• So x coordinate of B is -1 (∵ B is in the second quadrant)
(ii) In the right triangle OBD, the altitude will be √3 times the smallest side
• So BD = √3×OD = √3 units
• So distance of B from the x-axis is √3
• So y coordinate of B is √3
(iii) From (i) and (ii) we get: Coordinates of B are: (-1,√3)
5. Now we can calculate the length of the chord AB
AB = √[(-1-√3)2 + (√3-1)2] = √[(√3+1)2 + (√3-1)2] = √[3 + 2√3 +1 + 3 - 2√3 + 1]
= √[8] = √[4×2] = √4 × √2 = 2√2 units
In the next section we will learn about Tangents of circles.
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