Chapter 34.4 - Solved examples on Section formula

In the previous section we saw the section formula. We also saw some examples. In this section we will see a few more solved examples.

Solved example 34.6
The coordinates of the vertices of a quadrilateral are (2,1), (5,3), (8,7) and (4,9) in order.
(i) Find the coordinates of the midpoints of all sides
(ii) Prove that the quadrilateral with these midpoints as vertices is a parallelogram
Solution:
• Let us name the coordinates:
A(2,1), B(5,3), C(8,7), D(4,9)
1. Now we will calculate the midpoints:
(i) Midpoint of AB (Let us call it P):
[^(x1+x2)⁄₂,^(y1+y2)⁄₂] = [⁽²⁺⁵⁾⁄₂,⁽¹⁺³⁾⁄₂] = [⁷⁄₂,⁴⁄₂] = (3.5,2)
(ii) Midpoint of BC (Let us call it Q):
[⁽⁵⁺⁸⁾⁄₂,⁽³⁺⁷⁾⁄₂] = [¹³⁄₂,¹⁰⁄₂] = (6.5,5)
(iii) Midpoint of CD (Let us call it R):
[⁽⁸⁺⁴⁾⁄₂,⁽⁷⁺⁹⁾⁄₂] = [¹²⁄₂,¹⁶⁄₂] = (6,8)
(iv) Midpoint of DA (Let us call it S):
[⁽⁴⁺²⁾⁄₂,⁽⁹⁺¹⁾⁄₂] = [⁶⁄₂,¹⁰⁄₂] = (3,5)
2. We took the vertices A, B C and D in order.
• So P, Q, R and S are the midpoints taken in order:
P(3.5,2), Q(6.5,5), R(6,8), S(3,5)
• We have to prove that these points are vertices of a parallelogram
■ If both pairs of opposite sides are equal, then the quadrilateral is a parallelogram. 
3. In our case:
• One pair of opposite sides is [PQ, RS]
• The other pair of opposite sides is [QR,SP]
(A rough sketch of a quadrilateral will help to identify opposite sides)
4. We will use the distance formula to calculate the distances
• Length of PQ = √[(6.5-3.5)² + (5-2)²] = √[(3)² + (3)²] = √[2×(3)²] = 3√2
• Length of RS = √[(3-6)² + (5-8)²] = √[(-3)² + (-3)²] = √[2×(-3)²] = 3√2
• Length of QR = √[(6-6.5)² + (8-5)²] = √[(-0.5)² + (3)²] = √[0.25 + 9] = √9.25
• Length of SP = √[(3.5-3)² + (2-5)²] = √[(0.5)² + (-3)²] = √[0.25 + 9] = √9.25
■ Thus we find that PQ = RS AND QR = SP
So both pairs of opposite sides are equal. Thus it is a parallelogram
• The actual points in the Cartesian plane are shown in fig.34.13 below:

Fig.34.13

Solved example 34.7
In the fig.34.14(a) below, the midpoints of the sides of the large quadrilateral ABCD are joined to draw the small quadrilateral PQRS inside

Fig.34.14

(i) Find the coordinates of the fourth vertex P of the small quadrilateral
(ii) Find the coordinates of the other three vertices B, C and D of the large quadrilateral
Solution:
• We have seen in an earlier chapter that, the quadrilateral obtained by joining the midpoints of any other quadrilateral will always be a parallelogram
    ♦ See solved example 18.13 in section 18.7
• So PQRS is a parallelogram. We have seen the method to find the unknown vertex of a parallelogram when all the other three vertices are given
1.So the first step is to find the coordinates of P
(i) Group the vertices into two: [P,Q] and [R,S]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the second group are known. So we can write the details of the travel within that group:
(ii) To reach S from R:
• First travel 3 units horizontally to the left [∵ (3-6) = -3]
• Then travel 3 unit vertically downwards [∵ (3-6) = -3]
(iii) The same procedure of travel must be followed for the travel from Q to P
• First travel 3 units horizontally to the left
    ♦ At the end of this travel, the coordinates will be (6,5) [∵ (9-3) = 6]
• Then travel 3 unit vertically downwards
    ♦ At the end of this final lap, the coordinates will be (6,2) [∵ (5-3) = 2]
■ So the coordinates of P are (6,2)
2. P is the midpoint of AB
(i) Let the coordinates of A be (x1,y1)
    ♦ They are already given as (2,1)
(ii) Let the coordinates of B be (x2,y2)
(iii) Applying midpoint formula, we have:
• x coordinate of the midpoint P = ^(x1+x2)⁄₂ = ^(2+x2)⁄₂ = 6
⟹ (2+x2) = 12 ⟹ x2 = 10
• y coordinate of the midpoint P = ^(y1+y2)⁄₂ = ^(1+y2)⁄₂ = 2
⟹ (1+y2) = 4 ⟹ y2 = 3
(iv) Thus coordinates of B are (10,3)
3. Q is the midpoint of BC
(i) Let the coordinates of B be (x1,y1)
    ♦ They are already obtained as (10,3)
(ii) Let the coordinates of C be (x2,y2)
(iii) Applying midpoint formula, we have:
• x coordinate of the midpoint Q = ^(x1+x2)⁄₂ = ^(10+x2)⁄₂ = 9
⟹ (10+x2) = 18 ⟹ x2 = 8
• y coordinate of the midpoint Q = ^(y1+y2)⁄₂ = ^(3+y2)⁄₂ = 5
⟹ (3+y2) = 10 ⟹ y2 = 7
(iv) Thus coordinates of C are (8,7)
4. R is the midpoint of CD
(i) Let the coordinates of C be (x1,y1)
    ♦ They are already obtained as (8,7)
(ii) Let the coordinates of D be (x2,y2)
(iii) Applying midpoint formula, we have:
• x coordinate of the midpoint R = ^(x1+x2)⁄₂ = ^(8+x2)⁄₂ = 6
⟹ (8+x2) = 12 ⟹ x2 = 4
• y coordinate of the midpoint R = ^(y1+y2)⁄₂ = ^(7+y2)⁄₂ = 6
⟹ (7+y2) = 12 ⟹ y2 = 5
(iv) Thus coordinates of D are (4,5)
• The actual points in the Cartesian plane are shown in fig.34.14(b) above

Solved example 34.8
The vertices of a triangle ABC are points with coordinates (3,5), (9,13) and (10,6). Prove that it is an isosceles triangle. Calculate it's area.
Solution:
1. Fig.34.15(a) below shows a rough sketch

Fig.34.15

• We will use the distance formula to calculate the distances
• Length of AB = √[(9-3)² + (13-5)²] = √[(6)² + (8)²] = √[36 + 64] = √[100] = 10
• Length of BC = √[(10-9)² + (6-13)²] = √[(1)² + (7)²] = √[1 + 49] = √[50] = √[2×(5)²] = 5√2
• Length of AC = √[(10-3)² + (6-5)²] = √[(7)² + (1)²] = √[49 + 1] = √[50] = √[2×(5)²] = 5√2
2. So two sides AC and BC are equal. It is an isosceles triangle
• The meeting point of the two equal sides will be the apex. 
• And the third side will be the base
• So C is the apex and AB is the base
3. In an isosceles triangle, if we drop a perpendicular from the apex to the base, then the foot of the perpendicular will be the midpoint of the base. 
• Let this midpoint be D
• So we must find the coordinates of D. We can use the midpoint formula:
• x coordinate of the midpoint D = ^(x1+x2)⁄₂ = ⁽³⁺⁹⁾⁄₂ = 6
• y coordinate of the midpoint D = ^(y1+y2)⁄₂ = ⁽⁵⁺¹³⁾⁄₂ = 9
Thus the coordinates of D are: (6,9)
4. Now, CD is the altitude. We can use the distance formula to find it's length:
• Length of CD = √[(6-10)² + (9-6)²] = √[(-4)² + (3)²] = √[16 + 9] = √[25] = 5
5. Area of the triangle = ¹⁄₂ × base × altitude 
= ¹⁄₂ × AB × CD = ¹⁄₂ × 10 × 5 = 25 square units
• The fig.34.15(b) above shows the actual positions of the points in the Cartesian plane

Solved example 34.9
The coordinates of the vertices of a triangle are (-1,5), (3,7), (3,1). Find the coordinates of it's centroid
Solution:
Fig.34.16(a) below shows a rough sketch.

Fig.34.16

• D is the midpoint of AB. This midpoint is joined to the opposite vertex C. So CD is a median
• E is the midpoint of AC. This midpoint is joined to the opposite vertex B. So BE is a median
• The point of intersection of the two medians will give the centroid of the triangle ABC (Details here)
• Now we can write the steps:
1. Coordinates of D using the midpoint formula:
• x coordinate = ^(x1+x2)⁄₂ = ^(-1+3)⁄₂ = 1
• y coordinate = ^(y1+y2)⁄₂ = ⁽⁵⁺¹⁾⁄₂ = 3
2. Coordinates of E using the midpoint formula:
• x coordinate = ^(x1+x2)⁄₂ = ^(-1+3)⁄₂ = 1
• y coordinate = ^(y1+y2)⁄₂ = ⁽⁷⁺⁵⁾⁄₂ = 6
3. The centroid will divide the median in the ratio 2:1 measured from the vertex
• Let us divide the median CD in the ratio 2:1
    ♦ This ratio 2:1 is measured from the vertex. 
    ♦ So the segment corresponding to 2 will lie near the vertex C of the triangle
    ♦ And the segment corresponding to 1 will lie near the base side AB
4. We will use the section formula:
• m:n is 2:1
• k = 2+1 = 3
• The segment corresponding to 'm' lies near (x1,y1)
    ♦ So the segment corresponding to '2' lies near (x1,y1)
    ♦ Thus coordinates of C must be taken as (x1,y1) and coordinates of D must be taken as (x2,y2)  
5. Then x coordinate of the centroid = x1 + {[^m⁄_k]×(x2-x1)} 
= 3 + {[²⁄₃]×(1-3)} 
= 3 + {[²⁄₃]×(-2)} 
= 3 - ⁴⁄₃ = ⁵⁄₃
6. y coordinate of the centroid = y1 + {[^m⁄_k]×(y2-y1)} 
= 7 + {[²⁄₃]×(3-7)}
= 7 + {[²⁄₃]×(-4)} 
= 7 - ⁸⁄₃ = ¹³⁄₃
• So coordinates of the centroid are (⁵⁄₃,¹³⁄₃)
Check:
We can check the result by using the other median BE:
• Let us divide the median BE in the ratio 2:1
    ♦ This ratio 2:1 is measured from the vertex. 
    ♦ So the segment corresponding to 2 will lie near the vertex B of the triangle
    ♦ And the segment corresponding to 1 will lie near the base side AC
1. We will use the section formula:
• m:n is 2:1
• k = 2+1 = 3
• The segment corresponding to 'm' lies near (x1,y1)
    ♦ So the segment corresponding to '2' lies near (x1,y1)
    ♦ Thus coordinates of B must be taken as (x1,y1) and coordinates of E must be taken as (x2,y2)  
2. Then x coordinate of the centroid = x1 + {[^m⁄_k]×(x2-x1)} 
= 3 + {[²⁄₃]×(1-3)} 
= 3 + {[²⁄₃]×(-2)} 
= 3 - ⁴⁄₃ = ⁵⁄₃
3. y coordinate of the centroid = y1 + {[^m⁄_k]×(y2-y1)} 
= 1 + {[²⁄₃]×(6-1)}
= 1 + {[²⁄₃]×(5)} 
= 1 + ¹⁰⁄₃ = ¹³⁄₃
• So coordinates of the centroid are (⁵⁄₃,¹³⁄₃)
• These are the same coordinates that we obtained earlier
• The actual positions of the points in the Cartesian plane are shown in fig.34.16(b) above 

Solved example 34.10
The centre of a circle is (1,2) and (3,2) is a point on it. Find the coordinates of the other end of the diameter through this point
Solution:
1. Imagine a small horizontal line 
• Horizontal because y coordinates are same 
2. Put O(1,2) at it's center
3. Put the coordinates B(3,2) at the right end 
• B is towards right of O because x coordinate of B is greater than that of O
4. Let the left end be A
• We have to find the coordinates of A
5. AB is the diameter and O is the centre
• So OA = OB
6. Since OB is horizontal, it's length can be calculated from x coordinates alone. 
• That is., OB = (3-1) = 2 units
7. So OA = 2 units.
• So A is 2 units to the left of O
• Thus it's x coordinate will be (1-2) = -1
• y coordinate will be same as that of O and B, which is 2
• So the coordinates of A are (-1,2)

Solved example 34.11
Find the ratio in which the line segment joining the points (-3,10) and (6,-8) is divided by (-1,6)
Solution:
• Let A(x1,y1) = (-3,10)
• Let B(x2,y2) = (6,-8)
• Let P(-1,6) divide AB in the ratio m:n
1. Then x coordinate of P =  x1 + {[^m⁄_k]×(x2-x1)} = -1
⟹ -3 + {[^m⁄_k]×(6-(-3))} = -1
⟹ -3 + {[^m⁄_k]×(9)} = -1
⟹ {[^m⁄_k]×(9)} = 2 ⟹ {[^m⁄_k]} = ²⁄₉
⟹ {[^m⁄_(m+n)]} = ²⁄₉ ⟹ 9m = 2m + 2n ⟹ 7m = 2n
⟹ ^m⁄_n = ²⁄₇. That is., m:n = 2:7
Check:
Considering the y coordinates:
y coordinate of P =  y1 + {[^m⁄_k]×(y2-y1)} = 6
⟹ 10 + {[^m⁄_k]×(-8-10)} = 6
⟹ 10 + {[^m⁄_k]×(-18)} = 6
⟹ {[^m⁄_k]×(-18)} = -4 ⟹ {[^m⁄_k]} = ²⁄₉
⟹ {[^m⁄_(m+n)]} = ²⁄₉ ⟹ 9m = 2m + 2n ⟹ 7m = 2n
⟹ ^m⁄_n = ²⁄₇. That is., m:n = 2:7
• This is the same ratio as above.
• The actual positions of the points in the Cartesian plane are shown in the fig.34.17 below:

Fig.34.17

Another method of check:
1. Distance AB = √[(6-(-3))² + (-8-10)²] = √[(9)² + (-18)²] = √[81 + 324] = √[405] = √[5×9×9] = 9√5
2. Distance AP = √[(-1-(-3))² + (6-10)²] = √[(2)² + (-4)²] = √[4 + 16] = √[20] = √[5×2×2] = 2√5  
3. If AP is ²⁄₉ of AB, our calculations are correct
²⁄₉ of AB = ²⁄₉ × 9√5 = 2√5
4. From (2) we see that AP is indeed 2√5
• So our calculations are correct

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In the next section, we will see Lines.

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Chapter 31.7 - More Solved examples on Distance formula

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.

Fig.31.51

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:

Fig.31.52

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:

Fig.31.53

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:

Fig.31.54

(i) Draw a line at an angle of 30^o 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 30^o, 60^o 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 30^o)
• 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

• 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:

Fig.31.56

 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.

Fig.31.57

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 30^o, 60^o 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 30^o)
• 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 = 180^o
⟹ ∠DOB + 90 + 30 = 180^o ⟹ ∠DOB + 120 = 180^o ⟹ ∠DOB = 60^o
4. Then OAC is a 30^o, 60^o triangle.
(i) The hypotenuse will be 2 times the smallest side
• The smallest side is OD (∵ OD is opposite the smallest angle 30^o)
• 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

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In the next section we will learn about Tangents of circles.

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