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write the equation in standard form for the circle with center (10, 0) …

Question

write the equation in standard form for the circle with center (10, 0) passing through \\(\left(10, \frac{11}{2}\
ight)\\).

Explanation:

Step1: Recall circle standard form

The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

Step2: Identify center coordinates

Given center \((h, k) = (10, 0)\), so \(h = 10\), \(k = 0\).

Step3: Calculate radius

Radius \(r\) is the distance between center \((10, 0)\) and point \((10, \frac{11}{2})\) on the circle. Using distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), here \(x_1 = x_2 = 10\), \(y_1 = 0\), \(y_2=\frac{11}{2}\). So \(r=\sqrt{(10 - 10)^2+(\frac{11}{2}-0)^2}=\sqrt{0 + (\frac{11}{2})^2}=\frac{11}{2}\).

Step4: Substitute into standard form

Substitute \(h = 10\), \(k = 0\), \(r=\frac{11}{2}\) into \((x - h)^2 + (y - k)^2 = r^2\). We get \((x - 10)^2+(y - 0)^2 = (\frac{11}{2})^2\), which simplifies to \((x - 10)^2 + y^2=\frac{121}{4}\).

Answer:

\((x - 10)^2 + y^2=\frac{121}{4}\)