Find the equations of the tangents from..

Question

Find the equations of the tangents from the point $(0,1)$ to the circle $x^2+y^2-2x-6y+6=0$.

My Attempt:

Here, $$x^2+y^2-2x-6y+6=0$$ Comparing with $$x^2+y^2+2gx+2fy+c=0$$

Centre $(-g,-f)=(1,3)$ radius $r=\sqrt {g^2+f^2-c}=2units$.

Please help me to continue further.

Answer 1

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As you wrote, the center is $O(1,3)$ with radius $2$. Here, let $A(0,1),B(1,1)$.

It is easy to see that the line $AB$ : $\color{red}{y=1}$ is one of the tangent lines.

Now consider $\triangle{OAB}$. We get $$\tan\angle{OAB}=\frac{BO}{AB}=\frac{2}{1}=2$$ So, the slope of the other tangent line is given by $$\tan(2\angle{OAB})=\frac{2\cdot 2}{1-2^2}=-\frac{4}{3}$$ Hence, the equation of the other tangent line is $$y-1=-\frac 43(x-0)\iff \color{red}{y=-\frac 43x+1}$$

Answer 2

Using the implicit function theorem, it would be simple.

Consider $$F=x^2+y^2-2x-6y+6=0$$ Compute the partial derivatives $$F'_x=2x-2$$ $$F'_y=2y-6$$ and $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}= ????$$

Answer 3

Let the tangents be $y-1 = m(x-0)$. This must meet the circle in coincident points. Hence $$x^2 + (mx+1)^2 -2x - 6(mx+1) +6 = 0$$ must have equal roots. $$x^2(1+m^2) -2x(1+2m) + 1 = 0$$ has discriminant equal to zero. Thus $$(2m+1)^2 = (m^2+1) \Rightarrow m(3m+4) = 0$$ and hence the slopes of the tangents are $0, -\dfrac{4}{3}$. The tangents are given by $$ y-1 = -\frac{4}{3}x$$ and $$y - 1 = 0$$