Chord of Contact

Coordinate Geometry
OFFERED PRICE: Rs. 1,484
View Details

Get extra Rs. 148 off

USE CODE: SELF10

Chord of Contact

 

Let AP and AQ be tangents to circle from point P(x₁, y₁). Then equation of PQ is known as equation of chord of contact. 

                    

For circle x² + y² = a², it is 

x₁x + y₁y = a² 

For general circle, it is 

x₁x + y₁y + g(x₁ + x) + f(y₁ +y) + c = 0 



Note: 

1. It is also written as T = 0 

2. The equation of chord AB [A ≡ (R cos α, R sin α); B ≡ (R cos β, R sin β)] of the circle x² + y² = R² is given by 

x cos ((α + β )/2) + y sin ((α - β )/2) = a cos ((α - β )/2) 

3. If a line y = mx + c intersects the circle x² + y² = a2 in two distinct points A and B then length of intercept AB = 2√((a² (1+m² )-c²)/(1+m² )) 



Caution: 

The equation of a chord of contact and the equation of the tangent on a point of the circle and both given by T = 0. The difference is that while in the case of a tangent the point (x₁, y₁) lies on the circle. In the case of a chord of contact (x₁, y₁) lies outside the circle. 



Illustration: 

Write the equations of tangents to the circle x² + y² = 9 and having slope 2. 

Solution: 

Tangents with slope ‘m’ are given by y = mx + a √(1+m² ) i.e. y = 2x ± 3 √(1+4). 



Note: 

These are two parallel tangents to the circle at the end of the diameter. 



Illustration: 

Write the equation of tangents to the circle x² + y² = 25 at the point (3, 4)? 

Solution: 

Point (3, 4) lies on the circle. 

Required equation of the tangent is : 3x + 4y= 25 using 

x₁x + y₁y = a2, where (x₁, y₁) ≡ (3, 4) 



Illustration: 

Write the equation of normal to x² + y² = 25 at (3, 4)? 

Solution: 

Recall: 

A normal to the circle passes through the centre of the circle. 

Normal is: y – 0 = 4/3 (x – 0) (using two point form of straight line) i.e. 3y – 4x = 0

To read more, Buy study materials of Circles comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.