Find the equation of the circle passing through the point ( - 1, - 3) and having its centre at the point of intersection of the lines x – 2y = 4 and 2x + 5y + 1 = 0.
The intersection of the lines: x – 2y = 4 and 2x + 5y + 1 = 0.
is (2, - 1)
∴ This problem is same as solving a circle equation with centre and point on the circle given.
The general form of the equation of a circle is:
(x - h)2 + (y - k)2 = r2
Where, (h, k) is the centre of the circle.
r is the radius of the circle.
In this question we know that (h, k) = (2, - 1), so for determining the equation of the circle we need to determine the radius of the circle.
Since the circle passes through ( - 1, - 3), that pair of values for x and y must satisfy the equation and we have:
⇒ ( - 1 - 2)2 + ( - 3 - ( - 1))2 = r2
⇒ ( - 3)2 + ( - 2)2 = r2
⇒ r2 = 9 + 4 = 13
∴ r2 = 13
⇒ Equation of circle is:
(x - 2)2 + (y - ( - 1))2 = 13
⇒ (x - 2)2 + (y + 1)2 = 13
Ans:(x - 2)2 + (y + 1)2 = 13