MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
rohan bhatia Grade: 11
        
sir pls solve ths qutnthe circle x2+y2-6x-10y+c=0 does not touch or intersect the co-ordinate axes and the point(1,4) is inside the circle.find the set of values of c.
7 years ago

Answers : (2)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear rohan


x2+y2-6x-10y+c=0


(x-3)2 +(y-5)2 = 34-c


circle does not intersect the co ordinate axis so √(34-c) <3


  or 34-c<9


 or  c >25


 


and 1,4 lie inside the circle so redius of circle must be greater then distance between the point (1,4) and (3,5)


√(34-c)> √(3-1)2 +(5-4)2


34-c  >√5


c < 34- √5


so   25 < c  < 34- √5




 


Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
We are all IITians and here to help you in your IIT JEE & AIEEE preparation.

All the best.

Regards,
Askiitians Experts
Badiuddin


 

7 years ago
sid
22 Points
										But sir ,answer that is mentioned here in book is (25,29).….......…....................///.........................
										
5 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 3,000 off
USE CODE: CART20
Get extra R 280 off
USE CODE: CART20

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details