External & Internal Contacts of Circles

The two circles having centres at C1(x1, y1) and C2(x2, y2) and radii r1 and r2 respectively will

(a) intersect in two real and distinct points if and only if

|r1 − r2| < C1C2 < r1 + r2

(b) touch each other externally if and only if

C1C2 = r1 + r2 and their point of contact C is given by

C ≡ $\large (\frac{r_1 x_2 + r_2 x_1}{r_1 + r_2} , \frac{r_1 y_2 + r_2 y_1}{r_1 + r_2})$

(c) touch each other internally if and only if

C1C2 = |r1 − r2| and their point of contact C is given by

C ≡ $\large (\frac{r_1 x_2 – r_2 x_1}{r_1 – r_2} , \frac{r_1 y_2 – r_2 y_1}{r_1 – r_2})$

(d) One circle lies outside the other if

C1C2 > r1 + r2

(e) One circle is contained in the other if

C1C2 < |r1 − r2|

Illustration : Find the equation of a circle having the lines x2 + 2xy + 3x + 6y = 0 as its normal and having size just sufficient to contain the circle x(x − 4) + y(y − 3) = 0.

Solution: Pair of normals are

(x + 2y)(x + 3) = 0.

∴ Normals are x + 2y = 0.

Point of intersection of normals is the centre of required circle i.e. C1 (−3 , 3/2) and centre of given circle is C2(2 , 3/2) and radius

$\large r_2 = \sqrt{4 + \frac{9}{4}} = \frac{5}{2}$

Let r1 be the radius of required circle

⇒  r1 = |C1 − C2| + r2

$\large r_1 = \sqrt{(-3-2)^2 + (\frac{3}{2}-\frac{3}{2})^2} + \frac{5}{2} = \frac{15}{2}$

Hence equation of required circle is

x2 + y2 + 6x − 3y − 54 = 0.

Common Tangents to Two Circles
(a) The direct common tangents to two circles meet on the line joining centres C1 and C2 and divide it externally in the ratio of the radii.

Working Rule to find direct common tangents:

Step 1: Find the coordinates of centre C1, C2 and radii r1, r2 of the two given circles.

Step 2: Find the coordinates of the point, say P dividing C1C2 externally in the ratio r1 : r2 Let P ≡ (h, k).

Step 3: Write the equation of any line through P (h , k) i.e. y − k = m(x − h) ….(1)

Step 4: Find the two values of m, using the fact that the length of the perpendicular on (1) from the centre C1 of one circle is equal to its radius r1

Step 5: Substituting these values of ‘ m ‘ in (1), the equation of the two direct common tangents can be obtained.

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

Working Rule to find transverse common tangents:

All the steps except the 2nd step are the same as above. Here in the second step the point R (h, k) will divide C1C2 internally in the ratio r1 : r2

Notes:

∎ When two circles are real and non-intersecting, 4 common tangents can be drawn.

∎ When two circles touch each other externally, 3 common tangents can be drawn to the circles.

∎ When two circles intersect each other at two real and distinct points, two common tangents can be drawn to the circles.

∎ When two circles touch each other internally one common tangent can be drawn to the circles

Image of the circle by the line mirror

Let the circle be S ≡ x2 + y2 + 2gx + 2fy + c = 0 and the line be L = lx + my + n = 0.

The radius of the image circle will be the same as that of the given circle.

Let the centre of the image circle be (x1 , y1).

∴ slope of C1C2 × slope of line L = − 1 …..(1)

and midpoint of C1C2 lies on lx + my + n = 0

$\large l(\frac{x_1 – g}{2}) + m (\frac{y_1 -f}{2}) + n = 0$ ……(2)

solving (1) and (2), we get (x1 , y1)

=> Required image circle will be

$\large (x-x_1)^2 + (y-y_1)^2 = (\sqrt{g^2 + f^2-c})^2$

Illustration : 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 ≡ (1 , 2) and C2 ≡ (0 , 4)

Let r1 and r2 be the radii of the circles

=> r1 = √5 and r2 = 2√5

Also C1C2 = √(1 + 4) = √5

But r1 + r2 = 3√5 and r2 − r1

= √5 = C1C2

Hence the circles touch each other internally.

Illustration : Find the range of parameter ‘ a ‘ for which the variable line y = 2x + a lies between the circles x2 + y2 − 2x − 2y + 1 = 0 and x2 + y2 − 16x − 2y + 61 = 0 without intersecting or touching either circle.

Solution: The given circles are C1 : (x − 1)2 + (y − 1)2 = 1

and C2 : (x − 8)2 + (y − 1)2 = 4

The line y − 2x − a = 0 will lie between these circle if centre of the circles lie on opposite sides of the line, i.e. (1 −2 −a) (1 −16 − a) < 0

=> a ∈ (−15 , −1)

Line wouldn’t touch or intersect the circles if ,

$\large \frac{|1-2-a|}{\sqrt{5}} > 1 \; , \; \frac{|1-16-a|}{\sqrt{5}} > 2$

=> | 1 + a| > √5 , |15 + a| > 2√5

=> a > √5 −1 or a < − √5 −1 , a > 2√5 −15 or a < −2√5 −15

Hence common values of ‘ a ‘ are (2√5 −15, −√5 −1)

Illustration : Prove that x2 + y2 = a2 and (x − 2a)2 + y2 = a2 are two equal circles touching each other.

Solution: Given circles are

x2 + y2 = a2 …..(1)

and (x − 2a)2 + y2 = a2 …..(2)

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

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

Now , $\large AB = \sqrt{(0-2a)^2 + 0} = 2 a = r_1 + r_2$

Hence the two circles touch each other externally.

Also Read :

Equations of Circle in various forms
Parametric Equation of a Circle
Position of a Point w.r.t a Circle
Equation of Tangent & Normal on the Circle
Chord of Contact
Radical Axis of two circles
Equation of Family of Circles

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