describes a circle $\mathrm{h}$ the origin when
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A circle in the $x y$-plane has a diameter with endpoints $(-1,-3)$ and $(7,3)$. If the poo $(t(0, b))$ ) on the circle and $b> 0$, what is the value of?

Step 1 ：Given a circle in the xy-plane with a diameter having endpoints (-1,-3) and (7,3).

Step 2 ：The center of the circle is the midpoint of the diameter. The midpoint formula is \((\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\).

Step 3 ：Substituting the given points into the midpoint formula, we get \((\frac{-1+7}{2}, \frac{-3+3}{2})\) which simplifies to (3,0).

Step 4 ：The radius of the circle is half the length of the diameter. The length of the diameter is calculated using the distance formula \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).

Step 5 ：Substituting the given points into the distance formula, we get \(\sqrt{(7-(-1))^2 + (3-(-3))^2}\) which simplifies to 10. Therefore, the radius is half of this, which is 5.

Step 6 ：The equation of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where (h,k) is the center and r is the radius.

Step 7 ：Substituting the values we found into the equation of the circle, we get \((x-3)^2 + y^2 = 25\).

Step 8 ：\(\boxed{(x-3)^2 + y^2 = 25}\) is the equation of the circle.