Question

Write the equation in standard form for the circle with center (0,7) passing through (6,5/2

Answer 1

The equation in standard form for the circle is given by:

$(x-a)^2+(y-b)^2=r^2$

where the center is (a, b) and the radius is r.

Then, we can replace the point (6, 5/2) and the center (0, 7) in the equation to find r:

$(6-0)^2+(\frac{5}{2}-7)^2=r^2$

Simplifying the expressions inside the parenthesis:

$(6)^2+(\frac{5-7\times2}{2})^2=r^2$$36+(\frac{5-14}{2})^2=r^2$$36+(\frac{-9}{2})^2=r^2$$36+\frac{81}{4}=r^2$$\frac{225}{4}=r^2$

Then:

$r^2=\frac{225}{4}$

Finally, the equation is:

$(x-0)^2+(y-7)^2=\frac{225}{4}$

Answer:

$x^2+(y-7)^2=\frac{225}{4}$