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  • The equation of circle passing through origin with radius 5 unit and centerd at (3, k) isOption: 1 <img alt="x^2+y^2-6x-8y=0" src="https://learn.care

The equation of circle passing through origin with radius 5 unit and centerd at (3, k) is

Option: 1

x^2+y^2-6x-8y=0


Option: 2

x^2+y^2+6x+8y=0


Option: 3

x^2+y^2-6x+8y=0


Option: 4

both (a) and (c)


Answers (1)

best_answer

given radius = 5

to find equation of circle we need point "h"

Observe that triangle OMC is right angle triangle

use paythagoren theorem

\\a^2=h^2+k^2\\5^2=3^2+k^2\\k=\pm4

equation of circle when k =4

\\(x-3)^2+(y-4)^2=5^2\\x^2-6x+9+y^2-8x+16=25\\x^2+y^2-6x-8y=0

 

equation of circle when k =-4

\\(x-3)^2+(y+4)^2=5^2\\x^2-6x+9+y^2+8x+16=25\\x^2+y^2-6x+8y=0

Posted by

Divya Prakash Singh

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