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  • The maximum distance of the point (4, 4) from the circle <img alt="\mathrm{x^2+y^2-2 x-15=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&p

The maximum distance of the point (4, 4) from the circle \mathrm{x^2+y^2-2 x-15=0} is

 

Option: 1

10


Option: 2

9


Option: 3

5


Option: 4

None of these


Answers (1)

best_answer

The maximum distance will be along the normal to the circle through (4, 4). The centre of the circle is (1, 0). The radius of the circle is 4. Hence the maximum

\mathrm{\text { distance }=\sqrt{(4-1)^2+4^2}+4=5+4=9}

Posted by

Sanket Gandhi

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