Write the standard equation for the circle center (-6, 7) and radius r = 9

Solution:

The standard form of an equation of a circle with center (h, k), a point P(x, y) on the circle and radius r is given by the following expression:

\(\sqrt{(x-h)^{2} + (y-k)^{2}} = r\) (1)

Squaring on both sides we have:

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

In the given problem (h, k) is (-6, 7) and the radius r is 9.

Substituting the values in equation (1) we get:

\(\sqrt{(x-(-6))^{2} + (y-7)^{2}} = 9\)

\(\sqrt{(x+6)^{2} + (y-7)^{2}} = 9\)

Squaring both sides we get,

\((x+6)^{2} + (y-7)^{2} = 9^{2}\)

\(x^{2} + 6^{2} + 2(x)(6) + y^{2} + (-7)^{2} -(2)(y)(7) = 81\)

\(x^{2} + 36 + 12x + y^{2} + 49 -14y = 81\)

x^{2} + y^{2} + 12x -14y + 85 = 81

\(x^{2} + y^{2} + 12x -14y + 4 = 0\)

Or

\(x^{2} + y^{2} + 12x - 14y = -4\)

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Write the standard equation for the circle center (-6, 7) and radius r = 9

Summary:

The standard equation for the circle center (-6, 7) and radius r = 9 is \(x^{2} + y^{2} + 12x -14y + 4 = 0\)