Question

Find the equation of the circle which is circumscribed about the triangle whose vertices are A( - 2, 3), b(5, 2) and C(6, - 1). Find the centre and radius of this circle.

Answer 1

The general equation of a circle: (x - h)² + (y - k)² = r² …(i), where (h, k) is the centre and r is the radius.

Putting A( - 2, 3), B(5, 2) and c(6, - 1) in (i) we get

h² + k² + 4h - 6k + 13 = r² …(ii) 

h² + k² - 10h - 4k + 29 = r² …(iii)

and h² + k² - 12h + 2k + 37 = r² …(iv) 

subtracting (ii) from (iii) and also from (iv), 

- 14h + 2k + 16 = 0⟹ - 7h + k + 8 = 0 

- 16h + 8k + 24 = 0⟹ - 2h + k + 3 = 0 

Subtracting,

5h - 5 = 0 ⟹ h = 1 

k = - 1

Centre = (1, - 1)

Putting these values in (ii) we get, radius

= \(\sqrt{1+1+4+6+13}\) = \(\sqrt{25}\) = 5

Equation of the circle is 

(x - 1)² + {y - ( - 1)}² = 5² 

(x - 1)² + (y + 1)² = 25.