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## How to find common chord between parabola and circle??

3 years ago

A collection of circles is called as a system or family of circles. There can be various types of family of circles

The general equation of a circle is x

^{2}+ y^{2}+ 2gx + 2fy + c = 0. Since this equation involves three unknowns i.e. g, f and c so we need at least three conditions to get a unique circle.For example, if we are given two circles and we want to determine the third circle touching both of them. We shall need one more condition. Without the condition we get the equation of family of circles which satisfies the two given conditions. Imposition of a third condition will result in the equation representing a particular circle.

Let us now see some of the ways of finding the family of circles when certain conditions are given:

1. Family of circles having a fixed centre

This equation is given by (x – h)

^{2}+ (y – k)^{2}= r^{2}(Figure given below)

Since (h, k) is fixed, so the only parameter varying is r. This is one parameter family of circles, and is the equation of the family of concentric circles. Fixation of the radius will give a particular circle.

2. Equation of Family of circles passing through intersection of two circles S_{1}=0 and S_{2}=0.

The general equation of the family of circles passing through the intersection of S1 and S2 in given by S_{1}+ kS_{2}= 0, where k? -1. Here again we have one-parameter (k) equation of family of circles. The particular value of the parameter gives a unique circle.Caution:

If k = –1, we get equation of common chord i.e. straight line instead of circle.

Let S_{1}= x^{2}+ y^{2}+ 2g_{1}x + 2f_{1}y + c_{1}= 0S2 = x

^{2}+ y^{2}+ 2g_{2}x + 2f_{2}y + c_{2}= 0Since, point lies on both the circles,

?x

_{A}^{2}+ y_{A}^{2}+ 2g_{1}x_{A}+ 2f_{1}y_{A}+ c_{1}= 0So, x

_{A}^{2}+y_{A}^{2}+ 2g^{2}x_{A}+ 2f^{2}y_{A}+ c_{2}= 0? x

_{A}^{2}+ y_{A}^{2}+ 2g_{1}x_{A}+ 2f_{1}y_{A}+ c1 + ? (x_{A}^{2}+ y_{A}^{2}+ 2g_{1 }x_{A}+ 2f_{1}y_{A}+ c_{1}) = 0? Point A(xA, yA) lies on S

_{1}+ ? S_{2}= 0 ? ? ? RSimilarly point B(xB, yB) lies on S

_{1}+ ? S_{2}= 0 ? ? ? R? S1 + ? S2 = 0 is the family of circles through the intersection of S

_{1}= 0 and S_{2}= 0.3. Equation of a circle circumscribing a triangle whose sides are given by L

_{1}= 0, L_{2}= 0, L_{3}= 0.This equation is given by L

_{1}L_{2}+?L_{2}L_{3}+ µL_{3}L_{1}= 0, provided coefficeint of xy = 0 and coefficient of x^{2}= coefficient of y^{2}.The particular value of the parameter ? and µ gives a unique circle.

4. Family of circles touching the circle S = 0 and line L = 0 at their point of contact

The required family is given by the equation S + ?L = 0, where ? is the required family.5. Family of circle passing through two given points A(x

_{1}, x_{1}) and B(x_{2}, y_{2})? Requiredfamily of circlesis

6. Family of circles touching a given line L = 0 at a point (x_{1}, y_{1}) on the line is(x – x

_{1})^{2}+ (y – y_{1})^{2}+ ?L = 0, the particular value of the parameters ? gives a unique circle.The equation of the family of circles which touch the line y – y

_{1}= m (x – x_{1}) at (x_{1}, y_{1}) for any values of m is (x – x_{1})^{2}+ (y – y_{1})^{2}+ ?[(y – y_{1}) – m(x – x_{1})] = 0.Here, we can have two sub-cases according as whether the line is parallel to x -axis or y-axis.

If this line through (x

_{1}, y_{1}) is parallel to x-axis, then the equation of the family of circles touching it at (x_{1}, y_{1}) becomes (x – x_{1})^{2}+ (y – y_{1})^{2}+ K(y – y_{1}) = 0.And if the line through (x

_{1}, y_{1}) is parallel to x-axis, then the equation of the family of circles touching it at (x_{1}, y_{1}) becomes (x – x_{1})^{2}+ (y – y_{1})^{2}+ K(y – y_{1}) = 0.7. Equation of circle circumscribing a quadrilateral whose sides in order are represented by the lines

L_{1}= 0, L_{2}= 0, L_{3}= 0 and L_{4}= 0.The equation of the required family is given by

L_{1}L_{3}+ ?L_{2}L_{4}= 0, provided coefficient of x^{2}= coefficient of y^{2}and coefficient of xy = 0.## Common chord of two circles

The line joining the points of intersection of two circles is called the common chord. If the equations of the circles are

S

_{1}= x^{2}+ y^{2}+ 2g_{1}x + 2f_{1}y + c_{1}= 0

S_{2}= x^{2}+ y^{2}+ 2g_{2}x + 2f_{2}y + c_{2}= 0, then the equation of the common chord is S_{1}– S_{2}= 0.Hence, the equation is 2x (g

_{1}- g_{2}) + 2y (f_{1}-f_{2}) + c_{1}– c_{2}= 0.## Some key points:

The length of the common chord is 2v(r

_{1}^{2}– p_{1}^{2}) = 2v(r_{2}^{2}– p_{2}^{2}), where p1 and p2 are the length of perpendicular drawn from the centre to the chord.In the above equation of common chord, the coefficient of x

^{2}and y^{2}in both the equations should be equal.Two circles touch each other if the length of their common chord is zero.

The maximum length of the common chord is equal to the diameter of the smaller circle.

## Angle of intersection of two circles:

The angle of intersection between two circles S = 0 ans S’ = 0 is deifned as the angle between their tangents at their point of intersection.

If S = x

^{2}+ y^{2}+ 2g_{1}x + 2f_{1}y + c_{1}= 0and S“ = x

^{2}+ y^{2}+ 2g_{2}x + 2f_{2}y + c_{2}= 0, are two cricles with radii r_{1}, r_{2}and ‘d’ is the distance between their centres then the angle of intersection? between them is given bycos ? = (r

_{1}^{2}+ r_{2}^{2}– d^{2})/ 2r_{1}r_{2}or cos ? = 2(g_{1}g_{2}+ f_{1}f_{2}) – (c_{1}+ c_{2})/ 2v(g_{1}^{2}+ f_{1}^{2}– c_{1}) v(g_{2}^{2}+ f_{2}^{2}– c_{2})## Note:

The two circles are said to intersect orthogonally if the angle of intersection of the circles i.e., the angle between their tangents at the point of intersection is 90°.

The condition for the two circles to cut each other orthogonally is 2gg_{1}+ 2ff1 = c + c_{1}where (–g, –f) and (–g_{1}, –f_{1}) are the centres of the respective circles, S = 0 and S_{1}= 0.When two circles intersect orthogonally then the length of the tangent of one circle from the centre of other circle is equal to the radius of the other circle.

## External and Internal Contacts of Circles

A necessary and sufficient condition for the two circles to intersect at two distinct points is r

_{1}+ r_{2}> C_{1}C_{2}>| r_{1}- r_{2}|, where C_{1}, C_{2}are the centres and r_{1}, r_{2}be the radii of the two circles.If two circles with centres C

_{1}(x_{1}, y_{1}) and C_{2}(x_{2}, y_{1}) and radii r_{1}and r_{2}respectively, touch each other externally, C_{1}C_{2}= r_{1}+ r_{2}. Coordinates of the point of contact are A = ((r_{1}x_{2}+r_{2}x_{1})/(r_{1}+r_{2}),(r_{1}y_{2}+r_{2}y_{1})/(r_{1}+r_{2})).The circles touch each other internally if C

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

The circles touch each other internally if C_{1}C_{2}= r_{1}– r_{2}.

Coordinates of the point of contact are

T = ((r_{1}x_{2}- r_{2}x_{1})/(r_{1}- r_{2}),(r_{1}y_{2}- r_{2}y_{1})/(r_{1}-r_{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}= v_{5}and r_{2}= 2v_{5}. Also C_{1}C_{2}= v(1 + 4)= v_{5}.But r

_{1}+ r_{2}= 3v_{5}and r_{2}– r_{1}= v_{5}= C_{1}C_{2}.Hence the circles touch each other internally.

Illustration:

Find all the common tangents to the circles x

^{2}+ y^{2}– 2x – 6y + 9 = 0 and x^{2}+ y^{2}– 6x – 2y + 1 = 0.## Solution:

The centres and the radii of the circles are

C_{1}(1, 3) and r_{1}= v(1 + 9 - 9) = 1, C_{2}(–3, 1) and r_{2}= v(9 + 1 - 1) = 3,C

_{1}C_{2}= v20, r_{1}+ r_{2}= 4 = v16 and C_{1}C_{2}> r_{1}+ r_{2}.Sense the circles are non-intersecting. Thus there will be four common tangents.

Transverse common tangents are tangents drawn from the point P which divides C

_{1}C_{2}internally in the ratio 1 : 3.Direct common tangents are tangents drawn from the point Q which divides C

_{1}C_{2}externally in the ratio 1 : 3.Coordinates of P are ((1(-3)+3.1)/(1+3),(1.1+3.3)/(1+3)) i.e. (0,5/2) and coordinates of Q are (3, 4).

Transverse tangents are tangents through the point (0,5/2).

Any line through (0,5/2) is y – 5/2 = mx …… (1)

? mx – y + 5/2 = 0

Applying the usual condition of tangency to any of the circle, we get

(m.1 – 3 + 5/2 = 0)/v((m

^{2}+1) ) = 1 ? (m-1/2)^{2}= m^{2}+ 1? – m – 3/4 = 0 or 0.m

^{2}– m – 3/4 = 0.? m = –3/4 and 8 as coefficient of m

^{2}is zero.Therefore from (1),

(y-5/2)/x = m = 8 and -3/4 ? x = 0 is a tangent and 3x + 4y – 10 = 0 is another tangent.

Direct tangents are tangents drawn from the point Q(3, 4).

Now proceeding as for transverse tangents their equations are y = 4, 4x – 3y = 0.

## Illustration:

Find the equation of the circle described on the common chord of the circles x

^{2}+ y^{2}– 4x – 5 = 0 and x^{2}+ y^{2}+ 8y+ 7 = 0 as diameter## Solution:

## Equation of the common chord is S

_{1}– S_{2}= 0? x + 2y + 3 = 0

Equation of the circle through the two circles is S1 + ?S2 = 0

? x

^{2}+ y^{2}-4/(1+? ) x+8?y/(1+? )+(7? -5)/(1+? ) = 0.Its centre (2/(1+? ),-4/(1+? )) lies on x + 2y + 3 = 0

? 2/(1+? )-8/(1+? ) + 3 = 0 ? 2 – 8? + 3 + 3? = 0 ? ? = 1.

Hence the required circle is x

^{2}+ y^{2}– 2x + 4y + 1 = 0.## Illustration:

The line Ax + By + C = 0 cuts the circle x

^{2}+ y^{2}+ ax + by + c = 0 in P and Q. The Line A?x + B?y + C? = 0 cuts the circle x^{2}+ y^{2}+ a'x + b'y + c' = 0 in R and S. If P, Q, R, S are concyclic, prove that

## Solution:

The equation of the circle through the first line and the first circle, i.e. through P and Q is

x

^{2}+ y^{2}+ ax + by + c + ?1 (Ax + By + C) = 0. ........................ (1)The equation of the circle through R and S is

x

^{2}+ y^{2}+ a'x + b'y + c + ?2 (A'x + B'y + C') = 0.................... (2)Since P, Q, R, S are concyclic, (1) and (2) are identical.

? 1 = (a+?1 A)/(a'+?2 A' )=(b+?1 B)/(b'+?2 B' )=(c+?1 C)/(c'+?2 C')

? ?1A-?2A' + a - a' = 0,

?1B -?2B' + b - b' = 0,

? 1C-? 2C' + c - c' = 0.

Illustration:

Show that the circle passing through the origin and cutting the circles x

^{2}+ y^{2}- 2a1x - 2b1y + c = 0 and x^{2}+ y^{2}- 2a_{2}x - 2b_{2}y + c = 0 orthogonally is

Solution:

Let the equation of the circle passing through the origin be

x

^{2}+ y^{2}+ 2gx + 2fy = 0. ..................... (1)It cuts the given two circles orthogonally

? -2ga

_{1}- 2fb1 = c_{1}? c_{1}+ 2ga_{1}+ 2fb_{1 }= 0 ?_{2}and -2ga

_{2}- 2fb_{2}= c_{2}? c_{2}+ 2ga_{2}+ 2fb_{2 }= 0 ? (3)

Illustration:Find the equation of the circle touching the line 2x + 3y + 1 = – at the point (1, -1) and is orthogonal to the circle which has the line segment having end points (0, -1) and (-2, 3) as the diameter.(IIT JEE 2004).

## Solution:

The equation of the circle touching the line 2x + 3y + 1 = – at the point (1, -1) is

(x-1)

^{2}+ (y+1)^{2}+ ?(2x + 3y + 1) = 0This gives x

^{2}+ y^{2}+ 2x(?-1) + y(3?+2) + (?+2) = 0 …. (1)Now, this is orthogonal to the circle having end point of diameter (0, -1) and (-2, 3).

This yields x(x + 2) + (y + 1)(y - 3) = 0

i.e. x

^{2}+ y^{2}+ 2x -2y -3 = 0Hence, 2(? -2)/ 2. 1 + 2(3? + 2)/ 2 (-1) = ? + 2-3

This gives ? -3/2

hence, from equation (1), we have the equation of circle as 2x

^{2}+ 2y^{2}– 10x – 5y +1 =0.## Illustration:

Find a circle passing through the intersection of x

^{2}+ y^{2}- 4 = 0 and x^{2}+ y^{2}- 6x + 5 = 0 which passes through the point (2, 1)?## Solution:

Family of required circles is S

_{1}+ ? S_{2}= 0? (x

^{2}+ y^{2}- 4) + ? (x^{2}+ y2 - 6x + 5) = 0Since the required circle passes through the point (2, 1), the previous equation is satisfied for the point (2, 1)

? (4 + 1 - 4) + ? (4 + 1 - 12 + 5) = 0

1 - 2? = 0 ? ? =1/2

? Equation of the required circle is

(x

^{2}+ y^{2}- 4) + 1/2 (x^{2}+ 2y - 6x + 5) = 0? x

^{2}+ y^{2}- 2x - 1 = 0

3 years ago

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