Problem in Analytic Geometry! :D

Question

Help me! I've been stuck on this problem for almost a night and our exams will come up tomorrow :D

Find the equation of the circle that passes through the points of intersection of the circles x^2+y^2-4x+6y=12, x^2+y^2+4x-8y=28 and has its center on the y axis.

Thanks for responding!

Answer 1

The brute force way to do this problem is to find the points of intersection of the two circles (this labor intensive and error prone). Then find the midpoint of the segment joining the points of intersection. That segment is a chord in the target circle. The perpendicular bisector of that chord passes through the center of the target circle. The center is to be located on the y-axis. Find the equation of the perpendicular bisector, let x = 0 to find the y-coordinate of the center. THEN, determine the radius of the circle by finding the distance from the center to one of the points of intersection.

To find the points of intersection:

1) subtract one equation from the other (the x² and y² terms drop out). What is left is the equation of the line passing through the points of intersection.

2) Solve that equation for either x or y and substitute into the equation of one of the circles.

3) You will get a quadratic. Use quadratic formula to find coordinates of points of intersection.

4) Use midpoint formula to determine midpoint of that chord (M).

5) Find the slope of that chord (m).

6) The equation of the perpendicular bisector will by: y - M.y = m(x - M.x).

7) Since the center is on y-axis, let x = 0 in that equation to get y-coordinate of center. Center is O(0, ?).

8) Now you need the radius². Use the distance formula to determine distance from center to one of the points of intersection.

9) Finally write the equation. x² + (y - O.y)² = R²

Good luck with all that. Patience and perseverance required to reach the solution with this technique.

There is an easier way.

x² + y² - 4x + 6y - 12 + k(x² + y² + 4x - 8y - 28) = 0 is actually the equation of a circle passing through the two points of intersection, where k is any real number, k ≠ -1 (in this case you get the line passing through the two points),

You can distribute the k, rearrange terms, and come up other forms. From there you can locate the center and radius for any value of k.

To see more details visit this graph:

desmos.com/calculator/v58h2aistr

Use the slider on k to see many more circles that contain the two points. Turns out when k = 1, the center is at (0, 1/2).

Visit the last line of the graph to see the simplest form for the equation of the target circle.