Question

Show that the points (–3, 2), (2, –3) and `(1, 2sqrt(3))` lie on the circumference of that circle, whose centre is origin.

Answer 1

Given: Points `A(-3, 2), B(2, -3), C(1, 2sqrt(3))` and centre O(0, 0).

Step-wise calculation:

1. Equation for a circle with centre at origin: x² + y² = r².

2. For A(–3, 2):

OA² = (–3)² + 2² 

= 9 + 4 

= 13 

⇒ OA = `sqrt(13)`

3. For B(2, –3): 

OB² = 2² + (–3)²

= 4 + 9

= 13

⇒ OB = `sqrt(13)`

4. For `C(1, 2sqrt(3))`: 

`OC^2 = 1^2 + (2sqrt(3))^2`

= 1 + 4 × 3 

= 1 + 12

= 13

⇒ OC = `sqrt(13)`

OA = OB = OC = `sqrt(13)`, so all three points are at the same distance `sqrt(13)` from the origin and therefore lie on the circumference of the circle centred at the origin (radius = `sqrt(13)`).