A circle in the $xy$-plane has a diameter with endpoints $(-2,-3)$ and $(6,3)$. If the point $(b,0)$ lies on the circle and $b \gt 0$, what is the value of $b$?

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Given the endpoints of the diameter, we can find the center of the circle by finding the midpoint of the endpoints.

$$\text{center}=\left( \frac{-2+6}{2},\frac{-3+3}{2}\right)$$ $$\text{center}=(2,0)$$

We are asked to find the a point $(b,0)$, which is either directly left or right of the center $(2,0)$. We need to calculate this distance by finding the radius. The radius is half of the length of the diameter, which can be found with the distance formula or pythagorean theorem.

$$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$ $$d = \sqrt{(-2-6)^2+(-3-3)^2}$$ $$d=\sqrt{64+36}$$ $$d=\sqrt{100}$$ $$d=10$$ $$r=5$$

Therefore, the points 5 left and 5 right of the center are $(-3,0)$ and $(7,0)$. If $b\gt 0$, the correct value of $b$ is $\boxed{7}$.

Once we find the center and diameter through the processes above, we can write the standard equation of the circle.

$$(x-h)^2+(y-k)^2=r^2, \text{ where } (h,k) \text{ is the center and } r \text{ is the radius}$$ $$(x-2)^2+(y-0)^2=5^2$$ $$(x-2)^2+y^2=25$$

We can substitute $(b,0)$.

$$(b-2)^2+0^2=25$$ $$(b-2)^2=25$$ $$(b-2)=\pm \sqrt{25}$$ $$b-2=\pm 5$$ $$b=2+5 \text{ and } 2-5$$ $$b=\boxed{7} \text{ and} -3$$