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I tried to solve using a formula in my textbook, however it is just appropriate if the point is on the circle. I just realized that the point (30,10) is outside the circle. Please your help.

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$y-10=m(x-30)$ is an equation of the tangent line or $mx-y-30m+10=0$.

Thus, $\frac{|10m-10-30m+10|}{\sqrt{m^2+1}}=10$ or $m^2=\frac{1}{3}$ and the rest is smooth.

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Let $y=m(x-30)+10$ . We have $$(x-10)^2+(m(x-30))^2=100$$ as a result $$(m^2+1)x^2-(60m+20)x+900m=0$$ and $$\Delta=(60m+20)^2-3600m(m^2+1)=0$$

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  • $\begingroup$ $$400(3m+1)^2-3600m(m^2+1)=0$$ . thus $$9m^3-9m^2+3m-1=0$$ In other words $$9\left(m-\frac 13\right)^3-\frac 23=0$$ $\endgroup$ – Behrouz Maleki Feb 3 '17 at 13:48

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