Relative Position of Two Circles: Examples


This lesson will cover a few examples illustrating relative position of two circles. I’ll take up one example for each case, and won’t be showing all the calculations in much detail. Go back to the previous lesson for the concept explanation, and head over here in case you forgot how to find the centre or the radius of a circle.

Example 1 Determine the relative position of the two circles x2 + y2 = 1 and x2 + y2 – 6x – 8y + 16 = 0.

Solution Here, the centres of the two circles are C1(0, 0) and C2(3, 4) respectively, making C1C2 = 5. Next, the radii r1 and r2 are 1 and 3 respectively.

As C1C2 > r1 + r2, the two circles lie outside each other.

 

Example 2 Determine the relative position of the two circles x2 + y2 – 6x + 8 = 0 and x2 + y2 + 2x – 8 = 0.

Solution In this case, the centres of the two circles are C1(3, 0) and C2(-1, 0) respectively, which means C1C2 = 4. The radii of the circles, r1 and r2, are 1 and 3 respectively.

This makes C1C2 = r1 + r2, implying that the two circles lie touch externally.

 

Example 3 Determine the relative position of the two circles x2 + y2 = 16 and x2 + y2 – 8x – 6y = 0

Solution In this case, the centres of the two circles are C1(0, 0) and C2(4, 3) respectively, hence C1C2 = 5. The radii of the circles, r1 and r2, are 4 and 5 respectively.

Now, C1C2 < r1 + r2. But this could mean anything – intersecting, touching internally, or one lying inside the other. So, whenever you find C1C2 to be smaller r1 + r2, you must compare it with r1 – r2 as well.

Here C1C2 is also greater than r1 – r2, which means that the two circles intersect at two points.

 

Example 4 Determine the relative position of the two circles x2 + y2 = 16 and x2 + y2 – 6x + 8 = 0

Solution Here, C1 is (0, 0) and C2 is (3, 0), which makes C1C2 = 3. Further, r1 = 4 and r2 = 1. We can see that C1C2 = r1 – r2 , which means the circles touch internally.

 

Example 5 Determine the relative position of the two circles x2 + y2 = 25 and x2 + y2 – 2x – 4y + 4 = 0

Solution Here, C1 is (0, 0) and C2 is (1, 2), which makes C1C2 = \( \sqrt{5} \) . Further, r1 = 5 and r2 = 1. We can see that C1C2 < r1 + r2, but cannot conclude anything yet. We must compare C1C2 with r1 – r2 as well.

As C1C2 < r1 – r2, the circle with centre C2 lies completely inside the other.

 

And that should cover everything about relative position of two circles. In the next lesson, I’ll talk about common tangents to two circles.