How do you find the equation of the circle with center at (3, 2) and through the point (5, 4)?

Question

8708 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-18T21:31:04

The equation is $x^2+y^2-6x-4y+5=0$ .

The generic equation of a circle with center $(c_x, c_y)$ and radius $r$ is:

$(x-c_x)^2+(y-c_y)=r^2$

in our case

$(x-3)^2+(y-2)^2=r^2$

What is missing here is the radius, but we can find it substituting the point $(5,4)$

$(5-3)^2+(4-2)^2=r^2$

$2^2+2^2=r^2$

$4+4=r^2$

$r=sqrt(8)$

The final equation is

$(x-3)^2+(y-2)^2=8$ or, expanding the squares,

$x^2+y^2-6x-4y+5=0$ .