# Equation of a circle : Intersection of lines

• Nov 14th 2009, 11:48 PM
FailCalculus
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
• Nov 15th 2009, 12:06 AM
CaptainBlack
Quote:

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
• Nov 15th 2009, 12:22 AM
ukorov
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.
\$\displaystyle r^2 = (4 - 2)^2 + (-1 - 3)^2\$
\$\displaystyle r^2 = 8\$

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