Question -
Answer -
Let us find the pointsof 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 withcentre (-2, 1) and passing through the point (0, 0).
We know that theradius of the circle is the distance between the centre and any point on theradius. So, we find the radius of the circle.
So, the equation is (x– p)2 + (y – q)2 = r2
Where, p = -2, q = 1
(x + 2)2 +(y – 1)2 = r2 …. (1)
Equation (1) passesthrough (0, 0)
So, (0 + 2)2 +(0 – 1)2 = r2
4 + 1 = a2
5 = r2
r = √5
We know that theequation of the circle with centre (p, q) and having radius ‘r’ is given by: (x– p)2 + (y – q)2 = r2
By substitute thevalues in the above equation, we get
(x – (-2))2 +(y – 1)2 = (√5)2
(x + 2)2 +(y – 1)2 = 5
x2 +4x + 4 + y2 – 2y + 1 = 5
x2 + y2 +4x – 2y = 0
∴ The equation of thecircle is x2 + y2 + 4x – 2y = 0.