help with point of intersection

Question

 

I have a circle on a graph and point A is (-1, -2).

 

 

what is the point of intersection of ΘA, ΘB, and ΘC.

 

I had to write the standard equation form for ΘA.  I got (x+1)² + (y + 2)² = 9 = 3²

 

I had to write the standard equation form for a ΘB with center (2,2) and readius 4 and graph circle B.  I got (x-2)² + (y -2)² = 4² = 16

 

I had write the equation of the circle with the point (1,-3) is on a circle with center (-2,1).  Graph it as ΘC.  I got (x+ 2)² + (y-1)² = r²

 

thanks

Answer 1

Go ahead and graph circles A and B.

 

To graph A, graph the point (-1, -2). Then count 3 units left, right, up, and down, graphing the points (-4, -2), (2, -2), (-1, 1), and (-1, -5).

 

To graph B, graph the point (2, 2). Then count 4 units left, right, up, and down, graphing the points (-2, 2), (6, 2), (2, 6), and (2, -2).

 

 

 

Notice that the two circles clearly have the point (2, -2) in common. Perhaps this will be the point of intersection. To find out for sure, we'll need to graph C.

 

 

 

You know the center of C is (-2, 1), so graph that point. To figure out the radius, you'll need to find the distance from the center (-2, 1) to the point (1, -3). Use the distance formula:

 

√((-2 - 1)² + (1 - (-3))²) = √((-3)² + 4²) = √(9 + 16) = √(25) = 5

 

The radius is 5. That means, by the way, that the equation of circle C is (x + 2)² + (y - 1)² = 25.

 

Now, count 5 units left, right, up, and down from (-2, 1) to finish graphing circle C. You'll graph the points (-7, 1), (3, 1), (-2, 6), and (-2, -4).

 

 

 

It looks again like our point of intersection could be (2, -2). Plug that point into all three circle equations. If it successfully solves all three equations, then you've found your point!