Question

Write the equation in standard form for the circle with center (0, -10) passing through (9/2, -16)

Answer 1

x^2 + (y + 10)^2 = 225/4

Explanation:

To find the standard form of a circle with center (h, k) and radius r, the equation is:

(x - h)^2 + (y - k)^2 = r^2

We are given the circle's center as (0, -10) and a point on the circle as (9/2, -16). We can use this information to find the radius of the circle.

Radius (r) = Distance between center and point on the circle

r = sqrt[(0 - 9/2)^2 + (-10 + 16)^2]

r = sqrt[(81/4) + 36]

r = sqrt[(81 + 144)/4]

r = sqrt(225)/2

r = 15/2

Now we can substitute the values of (h, k, and r) into the standard form equation to get:

(x - 0)^2 + (y - (-10))^2 = (15/2)^2

Simplifying and multiplying out the terms, we get:

x^2 + (y + 10)^2 = 225/4

Therefore, the standard form of the circle is:

x^2 + (y + 10)^2 = 225/4

Answer 2

so we know the center and a point it passes through, so the distance from the center to that point on the circle is its radius

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{0}~,~\stackrel{y_1}{-10})\qquad (\stackrel{x_2}{(9)/(2)}~,~\stackrel{y_2}{-16})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ \stackrel{ radius }{r}=\sqrt{(~~(9)/(2) - 0~~)^2 + (~~-16 - (-10)~~)^2} \implies r=\sqrt{((9)/(2) )^2 + (-16 +10)^2} \\\\\\ r=\sqrt{( (9)/(2) )^2 + ( -6 )^2} \implies r=\sqrt{ (81)/(4) + 36 } \implies r=\sqrt{ \cfrac{225}{4} }\implies r=\cfrac{15}{2} \\\\[-0.35em] ~\dotfill`

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{0}{h}~~,~~\underset{-10}{k})}\qquad \stackrel{radius}{\underset{(15)/(2)}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - 0 ~~ )^2 ~~ + ~~ ( ~~ y-(-10) ~~ )^2~~ = ~~\left( (15)/(2) \right)^2\implies x^2+(y+10)^2=\cfrac{225}{4}`