Intersection of Circles

Video lesson · from 18:413 worked examples

One lesson video covers all of Circles, so it opens at 18:41 for this topic — not from the beginning.

Consider two circles with radii $r_1$ and $r_2$ with $r_1 > r_2$ .

Let $d$ be the distance between the centres of the two circles.

	The circles do not touch	The circles meet at one point only	The circles meet at two distinct points
External	$d > r_1 + r_2$	$d = r_1 + r_2$	$r_1 - r_2 < d < r_1 + r_2$
Internal	$d < r_1 - r_2$	$d = r_1 - r_2$

Compare the distance between centres with the radii: touching externally means $d = r_1 + r_2$ ; touching internally means $d = r_1 - r_2$ . Work out all three numbers before concluding.
Do not stop at the distance: the marks are for the comparison and the conclusion, so state which case applies and why.
Keep distances exact: rounding can make a genuine tangency look like a near miss.
Find the centres and radii correctly first: if either circle is in general form, extract $(-g,-f)$ and $\sqrt{g^2+f^2-c}$ before comparing anything.

Example 1

Circle P has centre (-4,-1) and radius 2 units, circle Q has equation $x^2+y^2-2x+6y+1=0$ .

Show that the circles P and Q do not touch.

Circle P: Centre = $(-4, -1)$ , $r_P = 2$

Circle Q: $x^2+y^2-2x+6y+1=0$

Centre: $2g = -2 \implies g = -1$ , $2f = 6 \implies f = 3$ . Centre = $(1, -3)$ .

Radius: $r_Q = \sqrt{(-1)^2 + 3^2 - 1} = \sqrt{1 + 9 - 1} = \sqrt{9} = 3$

Calculate distance $d$ between centres P and Q:

d = \sqrt{(1 - (-4))^2 + (-3 - (-1))^2} = \sqrt{5^2 + (-2)^2} = \sqrt{25+4} = \sqrt{29} \approx 5.39

Calculate $r_P + r_Q$ :

r_P + r_Q = 2 + 3 = 5

Since $d > r_P + r_Q$ ( $\sqrt{29} > 5$ ), the circles do not touch.

Example 2

Circle R has equation $x^2+y^2-2x-4y-4=0$ , and circle S has equation $(x-4)^2+(y-6)^2=4$ .

Show that the circles R and S touch externally.

Circle R: Centre = $(1, 2)$

Radius: $r_R = \sqrt{(-1)^2 + (-2)^2 - (-4)} = \sqrt{1 + 4 + 4} = 3$

Circle S: Centre = $(4, 6)$ , $r_S = \sqrt{4} = 2$

Calculate distance $d$ between centres R and S:

d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5

Calculate $r_R + r_S$ :

r_R + r_S = 3 + 2 = 5

Since $d = r_R + r_S$ ( $5 = 5$ ), the circles touch externally.

Example 3

Circle A has equation $(x-3)^2+(y+3)^2=64$ , and circle B has equation $(x+1)^2+y^2=9$ .

Show that the circles A and B touch internally.

Circle A: Centre = $(3, -3)$ , $r_A = \sqrt{64} = 8$

Circle B: Centre = $(-1, 0)$ , $r_B = \sqrt{9} = 3$

Calculate distance $d$ between centres A and B:

d = \sqrt{(-1-3)^2 + (0 - (-3))^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16+9} = 5

Calculate $r_A - r_B$ :

r_A - r_B = 8 - 3 = 5

Since $d = r_A - r_B$ ( $5 = 5$ ), the circles touch internally.