In the previous section we saw how to calculate the length of chords. In this section, we will see circles through '2 points' and '3 points'.
Consider any two points as
shown in fig.17.24(a) below.
 |
| Fig.17.24 |
• A line can pass through those two points. That line
will be unique. That is., there will be only one line which pass
through any two given points.
• Now consider any 3 points as shown in
fig.b. No line can pass through those 3 points. But if the 3 points
are in a straight line, then a line can pass through them as shown in fig.c.
• Next we consider circles. First
consider any two points A and B, as shown in fig.17.25(a) below:
 |
| Fig.17.25 |
A circle can pass through
any two points. But is there any definite rule? Let us analyse:
In fig.a, A and B are any two
points. If a circle pass through both A and B, then the line AB is a
chord of that circle. The perpendicular bisector of AB is XY. The centre of the circle will lie somewhere on XY. But there are infinite points on XY. For each of those points, there will be a circle. Let us draw a few such circles:
1. Take
any convenient point O1 on XY. With O1 as centre, draw a circle with
radius O1A. We can see that the circle pass through both A and B.
2. Take another convenient point O2. With O2 as centre, draw a circle with radius O2A. We can see that
the circle pass through both A and B. This is shown in fig.b
3. Take another convenient point O3. With O3 as centre, draw a circle with radius O3A. We can see that
the circle pass through both A and B. This is shown in fig.c
■ Why do all the circles that we
draw pass through both A and B?
• Because, ABO1, ABO2, ABO3 etc., form a series of isosceles triangles. O1, O2, O3 etc., lie on
the perpendicular bisector of the base AB, and so these centres are
at equal distances from both A and B.
• This is the very basis of the
theorem 1 that we saw at the beginning of this chapter.
• Thus we see that infinite
number of circles can be drawn passing through any two given points.
Next we consider 3
points. How many circles will pass through any given 3 points? Let us
analyse:
■ A, B and C are the given
points. We have to draw a circle passing through all the 3 of them.
• First consider A and B. If a
circle pass through A and B, AB will be a chord of that circle.
• X1Y1 is the perpendicular bisector of AB. We can draw infinite number of
circles that pass through both A and B by taking centres on X1Y1. One such circle is shown in green colour in fig.17.26(a) below:
 |
| Fig.17.26 |
• Like the green circle, we can draw any number of cirles by taking centres on X1Y1. But
there is no guarantee that any of those circles will pass through C
also.
• Now consider points B and C.
If a circle pass through B and C, BC will be a chord of that circle.
• X2Y2 is the perpendicular bisector of BC. We can draw infinite
number of circles that pass through both B and C, by taking centres on X2Y2. One such circle is shown in magenta colour in fig.17.26(a)
• But there is no guarantee that any of those circles will pass
through A also.
■ What about the point of
intersection 'O' of X1Y1 and X2Y2?
• 'O' will lie on both X1Y1 and X2Y2.
♦ The circle drawn with O as
centre will pass through A and B because O is on X1Y1
♦ The circle drawn with O as
centre will pass through B and C because O is on X2Y2
• So the circle drawn with O as
centre will pass through A, B and C.
• Such a circle is unique. That
is., there is only one circle that will pass through all 3 given
points.
• This circle is shown in red colour in fig.b.
• In the fig.b, we see that the 3 points
A, B and C lies on a circle. So AC is also a chord.
• When we join AC,
we get a complete triangle ABC. The circle with centre O, passes
through all the 3 vertices of the triangle. Such a circle is called
the Circumcircle of the triangle. It is shown in fig.c
■ The centre of the Circumcircle is called the Circumcentre of the triangle
■ The radius of the Circumcircle is called the Circumradius of the triangle
Now we will see some solved examples
Solved example 17.10
Draw 3 triangles, with length
of sides 4 cm and 5 cm, and the angle between them 60o, 90o and 120o.
Draw the circumcircle of each. Note how the position of the
circumcentre changes.
Solution:
■ The 3 triangles and their
circumcircles are shown in fig.17.27 below. The perpendicular bisectors are shown in blue colour.
 |
| Fig.17.27 |
• In fig.a, the angle 60o between
the given sides is an acute angle. The circumcentre lies inside the
triangle
• In fig.b, the angle 90o between
the given sides is a right angle. The circumcentre lies on the
hypotenuse
• In fig.c, the angle 120o between the given sides is an obtuse angle. The circumcentre lies
outside the triangle
Solved example 17.11
The equal sides of an
isosceles triangle are 8 cm long. The radius of it’s circumcircle
is 5 cm. Calculate the length of the third side.
Solution:
A rough sketch is shown in
fig.17.28 below. Note that it is only a rough sketch. Actual construction
is not necessary to solve this problem.
 |
| Fig.17.28 |
1. ABC is the isosceles
triangle with equal sides 8 cm. CD is the perpendicular bisector of
the base AB. It splits AB into two equal parts AD and BD. Let AD = BD
= x cm.
2. The perpendicular bisector
CD will pass through the centre ‘O’
3. Radius is given as 5 cm. So
OB = OC = 5 cm
4. Consider ⊿OBD. Applying
Pythagoras theorem, we get: OD = √[52-x2]
5. Consider ⊿CBD. Applying
Pythagoras theorem, we get: CD = √[82-x2]
6. But CD = 5 + OD.
Substituting for CD and OD from (5) and (4) we get:
7. √[82-x2]
= 5 + √[52-x2]
8. Squaring both sides of (7)
we get: 82-x2 = {5 + √[52-x2]}2
9. Use the identity [(a+b)2
= a2 + 2ab + b2] on the right side. We get:
10. 82-x2
= 25 + 2 × 5 × √[52-x2] + [52-x2] ⇒ 82-x2 = 25 + 10√[52-x2]
+ 52-x2
⇒ 10√[52-x2]
= 25 + 25 - 64 ⇒ 10√[52-x2]
= -14 ⇒ √[52-x2] = -1.4 ⇒ 52-x2
= -1.4 × -1.4 = 1.96
⇒ x2 = 25 – 1.96 = 23.04 ⇒ x = √23.04 = 4.8
11. So the third side = 2x = 2 × 4.8 = 9.6 cm
11. So the third side = 2x = 2 × 4.8 = 9.6 cm
Solved example 17.12
Find the relation between a
side and the circumradius of an equilateral triangle
Solution:
A rough sketch is shown in
fig.17.29 below. Note that it is only a rough sketch. Actual construction
is not necessary to solve this problem.
 |
| Fig.17.29 |
1. ABC is the equilateral
triangle. Let the side be ‘S’ cm. CD is the perpendicular
bisector of the base AB. It splits AB into two equal parts AD and BD.
So AD = BD = S⁄2 cm.
2. For ease of writing the
steps, we will put S⁄2 = x cm. Let the radius be ‘r’ cm
3. The perpendicular bisector
CD will pass through the centre ‘O’
4. Consider ⊿OBD. Applying
Pythagoras theorem, we get: OD = √[r2-x2]
5. Consider ⊿CBD. Applying
Pythagoras theorem, we get: CD = √[S2-x2]
6. But CD = r + OD.
Substituting for CD and OD from (5) and (4) we get:
7. √[S2-x2]
= r + √[r2-x2]
8. Squaring both sides of (7)
we get: S2-x2 = {r + √[r2-x2]}2
9. Use the identity [(a+b)2
= a2 + 2ab + b2] on the right side. We get:
10. S2-x2 = r2 + 2 × r × √[r2-x2] + [r2-x2] ⇒ S2-x2 = r2 + 2r√[r2-x2] + r2-x2
⇒ S2-2r2 = 2r√[r2-x2] ⇒ {S2-2r2}2 = 4r2[r2-x2] ⇒ S4 - 4S2r2 + 4r4 = 4r4 - 4r2x2
⇒ S4 - 4S2r2 = -4r2x2 ⇒ S4 - 4S2r2 = –4r2(S⁄2)2 [∵ from (2), x = S⁄2]
⇒ S4 - 4S2r2 = S2r2 ⇒ S4 = 3S2r2 ⇒ S2 = 3r2 ⇒ S = r√3
1.In the above fig.17.29, imagine that OD is extended downwards until it meet the circle. Then it will become a radius of the circle.
2. In any equilateral triangle, each side will be a perpendicular bisector of radius. We saw the details about this earlier based on fig.2.5.
3. So in the above fig.17.29, OD is r⁄2 and AB is perpendicular to OD. So we have a right triangle: ⊿OBD
4. Applying Pythagoras theorem, we get: BD = √[r2-(r⁄2)2]
5. Simplifying we get: BD = √[3r2⁄4] ⇒BD = (r⁄2)√3
6. So AB = 2 × BD = 2×(r⁄2)√3 = r√3
This completes our present discussion on circles. In the next section, we will see some more solved examples related to the topics that we discussed in this chapter in general.
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