Find the equation of a circle that has center at $(6,0)$ and that goes through the intersection point of $x^2+y^2-4x=0$ and $x = 3$?

Question

Find the equation that goes through the intersection point of $x^2+y^2-4x=0$ and $x=3$. The center of the circle is $(6,0)$.How I'm gonna find the intersection point of the circle and the line?

Answer 1

The equation of any circle passing through the intersection can be written as

$$x^2+y^2-4x+K(x-3)=0$$

Now use the fact: the center of the new circle is at $(6,0)$

Alternatively,

let the required equation of the new circle be $$(x-6)^2+(y-0)^2=r^2$$

Now if $(x_1,y_1);(x_2,y_2)$ are the intersections,

$$r^2=(x_1-6)^2+(y_1-0)^2=(x_2-6)^2+(y_2-0)^2$$