Click to Chat

1800-1023-196

+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
```
Let A be the center of the circle. x^2+y^2-2x-4y-20=0. Suppose that the tangents at the points B(1,7) & D(4,-2) on the circle meet at the point C. Find the the area of the quadrilateral ABCD.

```
6 years ago

```							solution:tangent at point B(1,7) :x + 7y -( x+1) - 2(y+7) -20 = 05y -35 = 0y = 7........................ (1)tangent at point D(4,-2) :4x - 2y - (x +4) - 2(y-2) - 20 = 03x -4y - 20 = 0 .................. (2)intersection of (1) & (2)point c = (16, 7)so vertex of quadrilateral are A(1,2) B(1,7)  C( 16,7) D (4,-2)Area of ABCD = Area of ABC + Area of ADC( because Area of ABC = Area of ADC)Area of ABCD = 2 *(Area of ABC)Area of ABC = 75/2 (ABC is a right angle triangle)so Area of ABCD = 75Thanks and Regards,Ajay verma,askIITians faculty,IIT HYDERABAD
```
6 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions