Finding the equation of a circle given one point and radius

Question

What is the equation of the circle which passes through the point $(0,2)$ with radius $4$ and whose center lies on the line $y = x$?

Answer 1

The equation of a circle with center $(a,b)$ and radius $r$ is $$(x-a)^2 + (y-b)^2 = r^2.$$ You know $r=4$, $a=b$ and that $(0,2)$ is a point on the circle. Can you figure out $a$ from that?

Answer 2

The equation of the circle around $(x_0,y_0)$ with radius $r$ is $$ (x-x_0)^2+(y-y_0)^2=r^2.$$ Now with $y_0=x_0$ and $r=4$, this becomes $$\tag1 (x-x_0)^2+(y-x_0)^2=16.$$ Since $(0,2)$ shall be on the circle, we knwo that $$ \tag2(0-x_0)^2+(2-x_0)^2=16$$ Use $(2)$ to compute all possible values of $x_0$. Each of these gives you a valid equation if you plug it into $(1)$.

Answer 3

So, the equation of the circle can be written as $(x-a)^2+(y-a)^2=4^2$

As the circle passes through $(0,2);$

$(0-a)^2+(2-a)^2=4^2$

Solve this for $a$

Answer 4

I'll only post a start since you have not shown your work. Recall that the equation of a circle with center $(h,k)$ and radius $r$ is $$ (x-h)^2+(y-k)^2 = r^2. $$We are given that $r=4$ and you can substitute $(x,y)=(0,2)$ to get $$ h^2 + (2-k)^2 = r^2 = 16. $$Finally, since $(h,k)$ is on the line $y=x$, we have $h=k$, so we have $$ h^2 + (2-h)^2 = 16. $$Now, you can solve for $h,k$.