# Thread: Equation of a circle : Intersection of lines

1. ## Equation of a circle : Intersection of lines

The problem is:

Find the equation of a circle given its center (2, -3) and pass through the intersection of lines x - 2y = 6 and x + 3y = 1

2. Originally Posted by FailCalculus
The problem is:

Find the equation of a circle given its center (2, -3) and pass through the intersection of lines x - 2y = 6 and x + 3y = 1
1. find the point of intersection of the lines (solve the simultaneous equations).

2. find the distance from the point of intersection found above to the point (2,-3), this is the radius.

3. You now have the centre and radius, so you can write down the equation.

If you have any problems let us know what they are and we will provide help specific to your difficulties.

CB

3. x - 2y = 6
x = 2y + 6 ...(1)
x + 3y = 1
x = 1 - 3y ...(2)

2y + 6 = 1 - 3y
x = 4 and y = -1
Hence (4, -1) is a point on the circle with centre (2, -3) and radius r.
$r^2 = (4 - 2)^2 + (-1 - 3)^2$
$r^2 = 8$

$(y + 3)^2 + (x - 2)^2 = 8$
$x^2 + y^2 - 4x + 6y + 5 = 0$