Find the equation of the circle whose centre is (2, - 3) and which passes through the intersection of the lines 3x + 2y = 11 and 2x + 3y = 4.
The intersection of the lines: 3x + 2y = 11 and 2x + 3y = 4
Is (5, - 2)
∴ 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, - 3), so for determining the equation of the circle we need to determine the radius of the circle.
Since the circle passes through (5, - 2), that pair of values for x and y must satisfy the equation and we have:
⇒ (5 - 2)2 + ( - 2 - ( - 3))2 = r2
⇒ 32 + 12 = r2
⇒ r2 = 9 + 1 = 10
∴ r2 = 10
⇒ Equation of circle is:
(x - 2)2 + (y - ( - 3))2 = 10
⇒ (x - 2)2 + (y + 3)2 = 10
Ans:(x - 2)2 + (y + 5)2 = 10