Find the equation of the circle with center at the origin and which passes through the point (3,4) ?

1 Answer
May 20, 2018

#x^2+y^2=25#

Explanation:

The general equation is #(x-a)^2+(y-b)^2=r^2# where#(a,b)# is the centre and the radius is #r#.

Our centre is (0,0) #=> (x-0)^2+(y-0)^2=r^2#

The circle passes through (3,4), if we make a right angle triangle with this point and the origin. The base would be 3 squares long, the height would be 4 squares long and the hypotenuse would be the radius.

Using Pythagoras, #3^2+4^2=h^2#

#9+16=h^2#

#h=sqrt25=5# as the hypotenuse equals the radius we get

#x^2+y^2=25#