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

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