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External and Internal Contact of Circles

Posted on - 07-02-2017

JEE Math Circle

IIT JEE

External and Internal Contacts of Circles

If two circles with centres C1(x1, y1) and C2(x2, y2) and radii r1 and r2 respectively, touch each other externally, C1C2 = r1 + r2. Coordinates of the point of contact are.

.

The circles touch each other internally if

C1C2 = r1- r2.

Coordinates of the point of contact are

.

Common Tangents to Two Circles :

(a) The direct common tangents to two circles meet on the line of centres and divide it externally in the ratio of the radii.

(b) The transverse common tangents also meet on the line of centres and divide it internally in the ratio of the radii.

Note :

  1. When two circles are real and non-intersecting, 4 common tangents can be drawn.
  2. When two circles touch each other externally, 3 common tangents can be drawn to
  3. the circles.
  4. When two circles intersect each other, two common tangents can be drawn to the circles. .
  5. When two circles touch each other internally 1 common tangent can be drawn to
    the circles.

Example 1

Examine whether the two circles x2+y2-2x-4y = 0 and x2+y2-8y-4=0 touch each other externally or internally.

Solution:

Let C1 and C2 be the centres of the circles.


&⇒
C1 º(1, 2) and C2 º (0, 4)
. Let r1 and r2 be the radii of the circles.


&⇒
r1 = Ö5 and r2 = 2Ö5

Also C1C2 = =

But r1 + r2 = 3Ö5 and r2 - r1 = Ö5 = C1C2 .

Hence the circles touch each other internally.

Example 2

Prove that x2 + y2 = a2 and (x – 2a)2 + y2 = a2 are two equal circles touching each other. Find the equation of circle (or circles) of the same radius touching both the circles.

Solution:

Given circles are

x2 + y2 = a2 …(1)

and (x – 2a)2 + y2 = a2 …(2)

Let A and B be the centres and r1 and r2 the radii of the two circles (1) and (2) respectively. Then.

A º (0, 0), B º (2a, 0), r1 = a, r2 = a

Now AB =

Hence the two circles touch each other externally.

Let the equation of the circle having same radius ‘a’ and touching the circles (1) and (2) be

(x – a)2 + (y – b)2 = a2 …(3)

Its centre C is (a, b) and radius r3 = a

Since circle (3) touches the circle (1),

AC = r1 + r3 = 2a. [Here AC > |r1 – r3| as r1– r3= a – a = 0].


&⇒
AC2 = 4a2
&⇒
a2 + b2 = 4a2 …(4)

Again since circle (3) touches the circle (2)

BC = r2 + r3
&⇒
BC2 = (r2 + r3)2


&⇒
(2a – a)2 + b2 = (a + a)2
&⇒
a2 + b2 – 4a a = 0


&⇒
4a2 – 4a a = 0 [from (4)]


&⇒
a = a and from (4), we have b = ± a.

Hence, the required circles are

(x – a)2 + ( y aÖ3)2 = a2

or x2 + y2 – 2ax 2Ö3ay + 3a2 = 0

 
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