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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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