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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.

pi-π
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3 Answers3

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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 $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}= ????$$

Claude Leibovici
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  • Could you please explain me about Implicit function theorems and derivates? I am not familiar with these terms. – pi-π Sep 17 '16 at 05:50
  • @user354073. So, don't worry ! You will learn it quite soon. It is simple way to compute the derivative $\frac{dy}{dx}$ when you handle an implicit function $F(x,y)=0$. Cheers. – Claude Leibovici Sep 17 '16 at 06:03
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$\qquad\qquad\qquad$enter image description here

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 $$y-1=-\frac 43(x-0)\iff \color{red}{y=-\frac 43x+1}$$

mathlove
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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 $$y - 1 = 0$$

  • What is meant by 'coincident point'? – pi-π Sep 17 '16 at 10:31
  • A line meets a circle in two points in general. If these two points are same, we call them 'coincident'. A tangent necessarily meets a curve in coincident points. –  Sep 17 '16 at 10:33
  • But if I am not mistaken, a tangent meets a circle at one point and a secant meets a circle at two different points, isn't it? I get confused. Please clarify – pi-π Sep 17 '16 at 10:35
  • That one point is a "double point". If we vary the secant through a point, when the other point of intersection coincides with this point, we have the tangent. Even the slope of the tangent is defined as the limiting value of the slope of the secant as the other point of intersection moves to the initial point. –  Sep 17 '16 at 10:39
  • What is limiting value? – pi-π Sep 17 '16 at 10:44
  • I think you have not yet encountered Calculus. Limits are defined there. –  Sep 17 '16 at 10:47
  • No, I have not. I am just a 10th grader. – pi-π Sep 17 '16 at 10:49
  • Wait till you meet this exciting subject! –  Sep 17 '16 at 10:51