Home/Question #3709
JEE Main 2020Medium

Q.The equation of the circle with centre (3,4) and radius 5 is

a
(x−3)2+(y−4)2=25(x-3)^2 + (y-4)^2 = 25
b
x2+y2−6x−8y−16=0x^2 + y^2 - 6x - 8y - 16 = 0
c
x2+y2−6x−8y+16=0x^2 + y^2 - 6x - 8y + 16 = 0
d
x2+y2−6x−8y=0x^2 + y^2 - 6x - 8y = 0

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

Unlock Detailed Solution

Register for Free

Already a member? Login here