INTERSECTION OF TWO CIRCLES

Let the equation of two circles be S₁ = x² +y² 2g₁x+2f₁y+c₁=0 and S₂ = x² +y²

2g₂x + 2f₂y + c₂ = 0 and let their centres are represented by O₁ and O₂ and their radii be r₁ and r₂ respectively.

Case I: when both the circles are non-intersecting and one lies outside the other

It the distance between the centres of two given circles is greater than the sum of their radii i.e. O₁O₂ > r₁ +r_2. Then both the circles will be non intersecting. In this case there exist four common tangents to these two circles. Two direct common tangents and two transverse common tangent.

Working Rule to find direct common tangent:

Step I: First find the point of intersection of direct common tangents say Q, which divides O₁ O₂ externally in r₁ : r₂

Step II: Write the equation of any line passing through Q ($\alpha$, $\beta$), i.e. y-$\beta$ = m (x-$\alpha$)…….(1)

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

Step IV: Substitutes these values of 'm' in (1), the equation of the two direct common tangents can be obtained.

Working Rule to find transverse common tangent:

To fine the equations of transverse common tangent first find the point of intersection of transverse common tangents say P, which divides O₁O₂ internally in r₁:r₂. Then follow the step 2, 3 and 4.

Case II: If the distance between the centres of the given circle is equal to sum of theirs radii. In this case both the circle will be touching each other externally. In this case two direct common tangents are real and distinct while the transverse tangents are coincident.

O₁ O₂ =|r₁+r₂|

The point of contact P can be find by using the fact that it divides O₁ O₂ internally in r₁ : r₂ .

Case III: It the distance between the centres of the given circles is equal to difference of their radii i.e. |O₁ O₂| = |r₁-r₂|, both the circles touches each other internally.

In this case point of contact divides O₁ O₂ externally in r₁ : r_2.

In this case only one common tangent exist.

Case IV: It the distance between the centres of two given circle is less then the sum of their radii but greater then the difference of their radii i.e.

|r₁ - r₂| < O₁ O₂ < r₁ + r₂, in this case both the circle will intersect at two real and distinct points.

In this case there exist two direct common tangents.

ANGLE OF INTERSECTION OF TWO CIRCLES

Angle of intersection of two circles is defined as the angle between the tangents at this point of intersection. Here angle $\angle$O₁PO₂ = $\pi$ - $\theta$

$\Rightarrow \cos \theta =\left|\frac{O_1{P}^2+O_2{P}^2-O_1{O}_2^2}{2O_{1}P{O}_{2}P}\right|$

$\Rightarrow \cos \theta =\left|\frac{2g_1{g}_2+2f_1{f}_{{}^2}-C_1{C}_2^{}}{2\sqrt{g_1^2+f_1^2-C_1}\sqrt{g_2^2+f_2^2-C_2}}\right|$

$\Rightarrow \cos \theta =\left|\frac{r_1^2+r_2^2-O_1{O}_2^2}{2r_1{r}_2}\right|$

ORTHOGONAL INTERSECTION OF TWO CIRCLES

The two circles are said to intersect orthogonally if the angle between the tangents at their point of intersection is 90⁰.

The condition for two circles S₁=O and S₂=O to cut each other orthogonally is $2g_1{g}_2^{}+2f_1{f}_2=c_1+c_2.$

Note: It two circle are intersecting orthogonally the tangent to one circle at the point of the intersection passes through the centre of other circle.

Case V: It the distance between the centres is less than the difference of their radii, i.e. |O₁ O₂| < |r₁-r₂|, in this case one circle will lie completely inside the other circle. Hence there will be no common tangent.

Illustration 1: If the circles x² + y² + 2ax + 2by = 0 and x² + y² + 2bx + 2cy = 0 touch each other, then show that b² = ac.

Solution I : Both the circles are touching each other. Hence distance between the centres should be equal to sum of radii. Centre are (–a, –b) and (–b, –c) and their radii are $\sqrt{a^2+b^2}\text{and}\sqrt{b^2+c^2}$ respectively

$\Rightarrow \sqrt{{(a-b)}^2+{(b-c)}^2}=\sqrt{a^2+b^2}+\sqrt{b^2+c^2}$

squaring both sides

$\Rightarrow$ b⁴ – 2acb² + a²c² = 0 $\Rightarrow$ (b² – ac)² = 0 $\Rightarrow$ b² = ac

Solution II : Clearly both the circles are passing through the origin. Hence they will be ouching each other at (0, 0). Hence at (0, 0) they will be having a common tangent.

$\Rightarrow$ 0.x + 0.y + a(x + 0) + b(y + 0) = 0 $\Rightarrow$ ax + by = 0 …(1)

$\Rightarrow$ 0.x + 0.y + b(x + 0) + c(y + 0) = 0 $\Rightarrow$ bx + cy = 0 …(2)

(1) and (2) identical

a/b = b/c $\Rightarrow$ b² = ac

LOCATION OF A CIRCLE IN RELATION TO A CIRCLE

Let S₁ $\equiv$ x² + y² + 2g₁x + 2f₁y+ c₁ = 0 and S₂ $\equiv$ x² + y² + 2g₂x + 2f₂y+ c₁ = 0 be two circles. Let D be the discriminant for the quadratic equation in x (or y) obtained by eliminating y (or x) from the two equations of the circle. Then

(i) they are two intersecting circles if D > 0

(ii) they are nonintersecting (no common points) if D < 0

(iii) they touch each other if D = 0

(iv) If D < 0, i.e., the circles are nonintersecting then

(a) S₁ = 0 is outside S₂ = 0 if S₂ (-g₁, –f₁) > 0 or S₁ (–g₂, –f₂) > 0; equivalently, AB > r₁ + r₂ where A, B are centres and r₁, r₂ are radii respectively.

(b) S₁ = 0 is inside S₂ = 0 if S₂ (–g₁, –f₁) < 0; equivalently, AB < |r₂ – r₁|

(v) If D = 0, i.e., then the circles touches each other

(c) externally if AB = r₁ + r₂

internally if AB = | r₁ – r₂|

CHORD OF CONTACT

From a point P(x₁, y₁) out side the circle two tangents PA and PB can be drawn to the circle. The chord AB joining the points of contact A and B of the tangents from P is called the chord of contact of P(x₁, y₁) with respect to the circle. Its equation is given by T = 0.

Illustration 2: If the chord of contact of the tangents drawn to x²+y²=a² from any point on x²+y²=b², touches the circle x²+y²=c², then show that a²=bc

Solution 1: Let P(x₁, y₁) be any point on x²+y²=b² i.e. x₁²+y₁²=b². Equation of corresponding chord of contact is xx₁+yy₁-a²=0. It touches x²+y²=c²

\begin{align}  \Rightarrow \dfrac{|-{{a}^{2}}|}{\sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}}}=|c| \\  \Rightarrow {{a}^{2}}=|bc| \\ \end{align}

Solution 2:

In $\Delta$OAC $\cos \theta =\frac{OC}{OA}=\frac{c}{a}$ …(1)

and in $\Delta$OAP $\cos \theta =\frac{a}{b}$ …(2)

from (1) and (2) a² = bc

Illustration 3: Tangents to the circle x² + y² = a² cut the circle x² + y² = 2a² at P and Q. Prove that tangents at P and Q to the circle x² + y² = 2a² intersect at right angles.

Solution: Equation of tangent at any point (a cosq, a sin q) to the circle x² + y² = a² is xcosq + ysin $\theta$ = a ………(1)

Let the point of intersection of the tangents at P and Q be (h, k). If the tangents at P and Q intersect at right angles, then locus of (h, k) will be director circle of

x² + y² = 2a² i.e. x² + y² = 4a². PQ is chord of contact of the circle x² + y² = 2a² w.r.t. the point (h, k) i.e. equation of PQ is hx + ky = 2a² ……(2)

(1) and (2) are same equation $\frac{\cos \theta }{h}=\frac{\sin \theta }{k}=\frac{a}{2a^2}=\frac{1}{2a}$

cos²q + sin² $\theta$ = 1 $\Rightarrow$ h² + k² = 4a²

$\therefore$ locus of (h,k) is x² + y² = 4a² which is director circle to x² + y² = 2a²

FAMILY OF CIRCLES

(i) If S $\equiv$ x² + y² + 2gx + 2fy + c = 0 and S' $\equiv$ x² + y² + 2g'x + 2f'y + c' = 0 are two intersecting circles, then family of circles passing through the point of intersection of S and S' is given by S + $\lambda$S' = 0, (where l is a parameter $\lambda$ $\ne$ –1).

(ii) If S $\equiv$ x² + y² + 2gx + 2fy + c = 0 is a circle which is intersected by the straight line L=lx+my+n=0 at two real and distinct points, then the equation of the family of circles passing through the point of intersection of circle and the given line is given by S + $\lambda$L = 0.(wherel is a parameter).

(iii) The equation of a family of circles passing through two given points (x₁, y₁) and (x₂, y₂) can be written in the form.

\left( {x - {x_1}} \right)\,\,\left( {x - {x_2}} \right)\,\, + \,\,\left( {y - {y_1}} \right)\,\,\left( {y - {y_2}} \right) + \lambda \left| {\begin{array}{*{20}{c}}xy1\\{{x_1}}{{y_1}}1\\{{x_2}}{{y_2}}1\end{array}} \right| = 0 (where l is a parameter).

(iv) The equation of the family of circles which touch the line y – y₁ = m(x – x₁) at (x₁, y₁) for any value of m is (x – x₁)² + (y – y₁)² + $\lambda$[(y – y₁) –m(x – x₁)] = 0. If m is infinite, the equation is (x – x₁)² + (y – y₁)² + l(x – x₁) = 0.

Illustration 4: The equation of the common chord of two circles is x+ y = 1. One of the circles has the ends of a diameter at the points (1 , -3) and (4, 1) and the other passes through the point (1 ,2). Find the equation of the two circles.

Solution: Let c₁ is the circle whose end point of a diameter are given

$\therefore$ c₁ $\equiv$ (x – 1) (x – 4) + (y +3) ( y – 1) = 0

$\equiv$ x² + y² – 5x + 2y + 1 = 0

Let c₂ is another circles

$\therefore$ Equation of family of circles passing through the points of intersection of circle c₁ and the given line

x² + y² – 5x + 2y +1 + l(x + y – 1) = 0 …………….(1)

for c₂ , 1^st equation satisfy the point (1 ,2 )

$\therefore$ $\lambda$= $-\frac{5}{2}$

$\therefore$ c₂ = 2x² + 2y² – 15x – y + 7 = 0

RADICAL AXIS

The radical axis of two circles is the locus of a point from which the tangent segments to the two circles are of equal length.

Equation to the Radical Axis

Consider S $\equiv$ x² + y² + 2gx + 2fy + c = 0

and S' $\equiv$ x² + y² + 2g'x + 2f'y + c' =0, then S-S'=0 represents the equation of the Radical Axis to the two circles i.e. 2x(g – g') + 2y(f – f') + c – c' = 0.

Note:

1. If S = 0 and S' = 0 intersect in real and distinct points then S – S' = 0 is the equation of the common chord of the two circles.

2. If S' = 0 and S = 0 touch each other, then S – S' = 0 is the equation of the common tangent to the two circles at the point of contact.

3. To find the equation of the radical axis of two circles, first make the coefficients of x² and y² in the equation of the two circles equal to unity.

4. The radical axis of two circles is perpendicular to the line joining their centres.

5. The radical axis of three circles taken in pairs meet at a point , called the radical centre of the circles. Coordinates of radical centre can be found by solving the equations S1=S2=S3=0.

6. The radical centre of three circles described on the sides of a triangle as diameters is the orthocentre of the triangle.

7. If two circles cut a third circle orthogonally, then the radical axis of the two circles pass through the centre of the third circle. or the locus of the centre of a circle cutting two given circles orthogonally is the radical axis of the two circles.

8. The radical axis of the two circles will bisect their common tangents.

Illustration 5: If the circle x²+y²+2a₁x+2b₁y+c₁=0 bisects the circumference of x²+y²+2a₂x+2b₂y+c₂=0, then show that 2a₂ (a₁ – a₂)x + 2b₂(b₁ – b₂)y + c₁ – c₂ = 0

Solution: Clearly the centre of second circle i.e., (–a₂, –b₂) should lie on the common chord of a circles i.e., on the line 2 (a₁ – a₂)x + 2(b₁ – b₂)y + c₁ – c₂ = 0

Hence 2a₂ (a₁ – a₂)x + 2b₂(b₁ – b₂)y + c₁ – c₂ = 0