Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Get extra Rs. 223 off
USE CODE: COUPON10

```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.
```
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
Get extra Rs. 2,385 off
USE CODE: COUPON10