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The equation of a circle is $y^2 + x^2 = 4$

Two tangents of this circle intersect at point $(10, -5)$, what is the equation of these tangents?

I've tried to work the point of where one of the tangents meets the circle as $(a, b)$ and substitute it into the equation of the circle, and the equation of the tangent - which is of course perpendicular to the equation of the radius that ends at point $(a, b)$ however that proved fruitless as far as I can tell, are there any other methods?

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HINT: set as $y=ax+b \, \, $ the equation of one of the tangents. Since you know that the line passes through point $(10,-5)$, you can transform the equation of the line to eliminate $b $ and to keep only the parameter $a $. Then, you have a system to solve to find the intersection points: so you can obtain a second degree equation to be solved for $x $. Take the determinant (it will be a second degree expression containing $a $ ) and find the two values of $a $ for which the determinant is zero. These will be the two slopes of the tangents. From this you can easily get the intercepts as well.

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  • $\begingroup$ Thanks for the answer, but I do not understand how b can be eliminated from the equation as the only new equation I can come up with is $-5 = 10a + b$ by substituting in that point. Could you point me in the right direction? $\endgroup$ – Ronikos Nov 16 '16 at 19:03
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    $\begingroup$ Since you have $b=-5-10a $, you can rewrite the equation of the line as $y=ax-5-10a $. Now equalize this with the circle equation. $\endgroup$ – Anatoly Nov 16 '16 at 19:41
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    $\begingroup$ After equalizing, solve for $x $, take the determinant and procede as described above. $\endgroup$ – Anatoly Nov 16 '16 at 20:35
  • $\begingroup$ I got the answer, thanks a lot $\endgroup$ – Ronikos Nov 16 '16 at 20:50
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Any straight line passing through $(10,-5)$ is of the form

$y-(-5)=m(x-10)$, or, $mx -y -5-10m=0$ where $m$ is the slope of the line.

If the line is tangent to the circle $x^2+y^2=2^2$ then the perpendicular distance from the center $(0,0)$ of the circle to the line is equal to radius of the circle. Hence, we have

$|\frac{-5-10m}{\sqrt{1^2+{-m}^2}}|=2$.

Solving these we get two values of $m$. Substitute those into the equation of the straight line and you have your two tangents drawn from the point $(10,-5)$.

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  • $\begingroup$ Not any straight line: the equation of the vertical line through this point doesn’t have this form. Fortunately, neither of the tangents in this problem is vertical. $\endgroup$ – amd Sep 19 '17 at 0:26
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Working in homogeneous coordinates, if we have the point $\mathbf p$ and matrix of the circle $C$, then the degenerate conic $\mathbf p_\times^TC^{\small\triangle}\mathbf p_\times$ consists of the tangents to the circle that pass through $\mathbf p$. In this expression, $C^{\small\triangle}$ is the adjugate of $C$ and $\mathbf p_\times$ is the “cross product matrix” of $\mathbf p$.

In this problem, $\mathbf p=(10,-5,1)^T$ and $C=\operatorname{diag}(1,1,-4)$, so $$\mathbf p_\times=\left[\begin{array}{r}0&-1&-5\\1&0&-1\\5&1&0\end{array}\right]$$ and $$A=\mathbf p_\times^TC^{\small\triangle}\mathbf p_\times=\left[\begin{array}{r}21&50&40 \\ 50&96&-20 \\ 40&-20&-500 \end{array}\right].$$ This matrix is singular, so it does indeed represent a degenerate conic, and its null vector is $(10,-5,1)$, the point from which we’re drawing the tangents, so it checks out so far.

This matrix corresponds to the Cartesian equation $$21x^2+100xy+96y^2+80x-40y-500=0,$$ which is an equation of two intersecting lines, but you likely want the individual equations. You can try to factor the equation, but there’s a mechanical way to “split” the degenerate matrix. Since it represents a pair of lines, it is of rank 2 and has the form $\mathbf g\mathbf h^T+\mathbf h\mathbf g^T$. We proceed by finding a vector $\mathbf q$ and scalar $\alpha$ such that $A+\alpha\mathbf q_\times$ is a multiple of the rank-1 matrix $\mathbf g\mathbf h^T$, from which we can read $\mathbf g$ and $\mathbf h$, the two lines that we’re looking for.

Form the adjugate of $A$, $$B=\left[\begin{array}{r}48400&-24200&4840\\-24200&12100&-2420\\4840&-2420&484\end{array}\right]$$ (I’ve multiplied through by $-1$ here) and choose any non-zero diagonal element. The bottom left element is recognizable as $2^2\cdot11^2$, so we’ll take that one. We then have for $\alpha\mathbf q$ the corresponding column of $B$ divided by the square root of this element: $(220,-110,22)^T$. Form the cross product matrix of this vector and add it to $A$, getting $$\left[\begin{array}{r}21&28&-70\\72&96&-240\\150&200&-500\end{array}\right].$$ Any row and column of this matrix that contain a non-zero diagonal element give us our two lines. After eliminating common factors from the resulting equations to make the coefficients smaller, we have for the equations of the tangents through $(10,-5)$ $$3x+4y=10 \\ 7x+24y=-50.$$

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