Question

Find the co-ordinates of the circumcenter of the triangle whose vertices are (3,
7), (0,6) and (-1,5). Find the circumradius.
(HOTS)
​

Answer 1

• Circumcenter is at (3, 2)

• Curcumradius is 5 units

Explanation:

We consider the given points as

• A (3, 7)

• B (0, 6)

• C (- 1, 5)

Step 1. [ For the line AB ]

The mid-point of AB is at

( (3 + 0)/2, (7 + 6)/2 ) i.e., (3/2, 13/2)

Slope of AB is

= (7 - 6)/(3 - 0) = 1/3

∴ the slope of the line perpendicular to AB is (- 3)

Now the equation of the perpendicular line of AB through the point (3/2, 13/2) is

y - 13/2 = - 3 * (x - 3/2)

or, 2y - 13 = - 6x + 9

or, 6x + 2y = 22

or, 3x + y = 11 ..... (1)

Step 2. [ For the line AC ]

The mid-point of AC is at

( (3 - 1)/2, (7 + 5)/2 ) i.e., (1, 6)

Slope of AC is

(7 - 5)/(3 + 1) = 1/2

∴ the slope of the line perpendicular to AC is (- 2)

Now the equation of the perpendicular line of AC through the point (1, 6) is

y - 6 = - 2 * (x - 1)

or, y - 6 = - 2x + 2

or, 2x + y = 8 ..... (2)

Step 3. [ Finding the intersection ]

The bisectors are

3x + y = 11

2x + y = 8

Solving, we get

x = 3 and y = 2

Therefore the required circumcenter is at (3, 2)

Step 4. [ Finding the curcumradius ]

The circumcenter is at (3, 2) and B (0, 6) is a vertex of the triangle.

Therefore the length of the curcumradius is

= √{(3 - 0)² + (2 - 6)²} units

= √(9 + 16) units

= √25 units

= 5 units