Find the equation of the 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 the lines x + y + 1 = 0 and x – 2y + 4 = 0.
Given that the circle has the centre at the intersection point of the lines x + y + 1 = 0 and x - 2y + 4 = 0 and passes through the point of intersection of the lines x + 3y = 0 and 2x - 7y = 0.
Let us find the points of intersection of the lines.
On solving the lines x + 3y = 0 and 2x - 7y = 0, we get the point of intersection to be (0, 0)
On solving the lines x + y + 1 and x - 2y + 4 = 0, we get the point of intersection to be ( - 2, 1)
We have circle with centre ( - 2,1) and passing through the point (0,0).
We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.
We know that the distance between the two points (x1,y1) and (x2,y2) is 
.
Let us assume r is the radius of the circle.
⇒ 
⇒ 
⇒ 
⇒ r = √5 ..... (1)
We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:
⇒ (x - p)2 + (y - q)2 = r2
Now we substitute the corresponding values in the equation:
⇒ (x - ( - 2))2 + (y - 1)2 = (
)2
⇒ (x + 2)2 + (y - 1)2 = 5
⇒ x2 + 4x + 4 + y2 - 2y + 1 = 5
⇒ x2 + y2 + 4x - 2y = 0.
∴The equation of the circle is x2 + y2 + 4x - 2y = 0.