The two circles having centres at C_{1}(x_{1}, y_{1}) and C_{2}(x_{2}, y_{2}) and radii r_{1} and r_{2} respectively will

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

|r_{1} − r_{2}| < C_{1}C_{2} < r_{1} + r_{2}

(b) touch each other externally if and only if

C_{1}C_{2} = r_{1} + r_{2} 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

C_{1}C_{2} = |r_{1} − r_{2}| 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

C_{1}C_{2} > r_{1} + r_{2}

(e) One circle is contained in the other if

C_{1}C_{2} < |r_{1} − r_{2}|

Illustration : Find the equation of a circle having the lines x^{2} + 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. C_{1} (−3 , 3/2) and centre of given circle is C_{2}(2 , 3/2) and radius

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

Let r_{1} be the radius of required circle

⇒ r_{1} = |C_{1} − C_{2}| + r_{2}

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

Hence equation of required circle is

x^{2} + y^{2} + 6x − 3y − 54 = 0.

Common Tangents to Two Circles

(a) The direct common tangents to two circles meet on the line joining centres C_{1} and C_{2} and divide it externally in the ratio of the radii.

__Working Rule to find direct common tangents:__

Step 1: Find the coordinates of centre C_{1}, C_{2} and radii r_{1}, r_{2} of the two given circles.

Step 2: Find the coordinates of the point, say P dividing C_{1}C_{2} externally in the ratio r_{1} : r_{2} 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 C_{1} of one circle is equal to its radius r_{1}

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 C_{1}C_{2} internally in the ratio r_{1} : r_{2}

__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 ≡ x^{2} + y^{2} + 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 (x_{1} , y_{1}).

∴ slope of C_{1}C_{2} × slope of line L = − 1 …..(1)

and midpoint of C_{1}C_{2} 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 (x_{1} , y_{1})

=> 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 x^{2} + y^{2} − 2x − 4y = 0 and x^{2} + y^{2} − 8y − 4 = 0 touch each other externally or internally.

Solution: Let C_{1} and C_{2} be the centres of the circles.

=> C_{1} ≡ (1 , 2) and C_{2} ≡ (0 , 4)

Let r_{1} and r_{2} be the radii of the circles

=> r_{1} = √5 and r_{2} = 2√5

Also C_{1}C_{2} = √(1 + 4) = √5

But r_{1} + r_{2} = 3√5 and r_{2} − r_{1}

= √5 = C_{1}C_{2}

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 x^{2} + y^{2} − 2x − 2y + 1 = 0 and x^{2} + y^{2} − 16x − 2y + 61 = 0 without intersecting or touching either circle.

Solution: The given circles are C_{1} : (x − 1)^{2} + (y − 1)^{2} = 1

and C_{2} : (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 x^{2} + y^{2} = a^{2} and (x − 2a)^{2} + y^{2} = a^{2} are two equal circles touching each other.

Solution: Given circles are

x^{2} + y^{2} = a^{2} …..(1)

and (x − 2a)^{2} + y^{2} = a^{2} …..(2)

Let A and B be the centre and r_{1} and r_{2} the radii of the two circles (1) and (2) respectively. Then

A ≡ (0 , 0), B ≡ (2a , 0), r_{1} = a, r_{2} = a

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

Hence the two circles touch each other externally.