CIRCLES**

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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.

Badiuddin askIITians.ismu Expert · Accepted Answer

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) 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

Analytical Geometry> CIRCLES**...

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.

rohan bhatia , 15 Years ago

Badiuddin askIITians.ismu Expert

sid

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Analytical Geometry> CIRCLES**...

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.

rohan bhatia , 15 Years ago

Badiuddin askIITians.ismu Expert

sid

LIVE ONLINE CLASSES

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