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write the equation in standard form of a circle with center (-7,6), tan…

Question

write the equation in standard form of a circle with center (-7,6), tangent to the y - axis. the standard form of the equation for of the circle is \boxed{}. (type an equation. simplify your answer)

Explanation:

Step1: Recall the standard form of a circle's equation

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

Step2: Determine the radius of the circle

The circle is tangent to the \(y\)-axis. The distance from the center \((-7, 6)\) to the \(y\)-axis is the absolute value of the \(x\)-coordinate of the center. The \(x\)-coordinate of the center is \(-7\), so the radius \(r = | - 7| = 7\).

Step3: Substitute the center and radius into the standard form

The center \((h, k)\) is \((-7, 6)\) and \(r = 7\). Substituting these values into the standard form equation \((x - h)^2 + (y - k)^2 = r^2\), we get \((x - (-7))^2 + (y - 6)^2 = 7^2\).
Simplifying \((x - (-7))\) gives \((x + 7)\), and \(7^2 = 49\). So the equation becomes \((x + 7)^2 + (y - 6)^2 = 49\).

Answer:

\((x + 7)^2 + (y - 6)^2 = 49\)