Question

Find the equation of circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of lines x + y + 1 = 0 and x – 2y + 4 = 0.

Answer 1

Required circle passes through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0. 

x + 3y = 0 

⇒ x = -3y ……..(i) 

2x – 7y = 0 ……(ii) 

Substituting x = -3y in (ii), we get 

⇒ 2(-3y) – 7y = 0

⇒ -6y – 7y = 0 

⇒ -13y = 0 

⇒ y = 0

Substituting y = 0 in (i), we get x = -3(0) = 0 

Point of intersection is O(0, 0). 

This point O(0, 0) lies on the circle. 

Let C(h, k) be the centre of the required circle. 

Since, point of intersection of lines x + y = -1 and x – 2y = -4 is the centre of circle. 

∴ x = h, y = k 

∴ Equations of lines become 

h + k = -1 ……(iii) 

h – 2k = -4 …..(iv) 

By (iii) – (iv), we get 3k = 3 

⇒ k = 1 

Substituting k = 1 in (iii), we get h + 1 = -1 

⇒ h = -2 

∴ Centre of the circle is C(-2, 1) and it passes through point O(0, 0). Radius(r) = OC

The equation of a circle with centre at (h, k) and radius r is given by 

(x – h)² + (y – k)² = r² 

Here, h = -2, k = 1 

the required equation of the circle is 

(x + 2)² + (y – 1)² = (√5)² 

⇒ x² + 4x + 4 + y² – 2y + 1 = 5 

⇒ x² + y² + 4x – 2y = 0