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 Question from Novelyn, a student: find an equation of the line tangent to a circle with equation x^2+y^2+6x-8y-27=0 at the point P(1,-2)

Hi Novelyn,

With the equation of the circle, $(x^2 + 6x) + (y^2 -8 y) =27$ complete the square for each of $x^2 + 6x$ and $y^2 -8 y$ to write the equation in the form $(x - h)^2 + (y - k)^2 = r^2.$ This tells you that the center of the circle is $(h, k).$ The line segment joining $(h, k)$ and $(1, -2)$ is a radius of the circle. Find its slope $m.$ The tangent to the circle through $(1, -2)$ is the line through $(1, -2)$ with slope $\large \frac{-1}{m}.$

Penny

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