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The equation of a circle with center (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2.
At any point (a, b) on a circle, the tangent is perpendicular to the line drawn from the point to the center.
The circle with center C has an equation (x-5)^2 + (y+12)^2 = 169. The center C is (5, -12). The equation of the tangent drawn through the point (-7, -7) that lies on the circle has to be determined.
The slope of AC is (-12 + 7)/(5 + 7) = -5/12
As the tangent is perpendicular to AC it has a slope 12/5
This gives the equation of the tangent as (y + 7)/(x + 7) = 12/5
=> 5y + 35 = 12x + 84
=> 12x - 5y - 49 = 0
The equation of the tangent to the circle (x-5)^2 + (y+12)^2 = 169 at the point (-7, -7) is 12x - 5y - 49 = 0
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