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Find the equations of tangents drawn from point $(11,3)$ to the circle $x^2+y^2=65$

How are we supposed to draw two tangents at a given point?

The answer is $7x-4y-65=0$ and $4x+7y-65=0$.

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  • $\begingroup$ Picture the unit circle and the point $(0,2)$. You could draw two lines through this point, one which is tangent to a point on the upper-half circle, and one which is tangent to a point on the lower-half circle. The same idea holds in this case. $\endgroup$ – infinitylord Dec 13 '17 at 2:01
  • $\begingroup$ Differentiate thru the entire equation. $\endgroup$ – Karn Watcharasupat Dec 13 '17 at 2:03
  • $\begingroup$ @ Karn Watcharasupat, with respect to what??? $\endgroup$ – pi-π Dec 13 '17 at 2:06
  • $\begingroup$ You want dy/dx, so do an implicit derivative in terms of x and solve for dy/dx $\endgroup$ – Kaynex Dec 13 '17 at 2:09
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The green circle has a diameter from the origin to (11,3). It turns out to be $$ (2x-11)^2 + (2y-3)^2 = 130.$$ A triangle inscribed in a circle (the green one) with one edge a diameter is actually a right triangle. It is now easy to confirm that the intersection points of the two circles are $(4,7)$ and $(7,-4).$

Oh, that is a square produced by the two right triangles I constructed. That is unusual. The square happened precisely because $11^2 + 3^2 = 2 \cdot 65 \; .$ Most of the time, the two right triangles would make a kite shape, the diagonals of the resulting quadrilateral would still be orthogonal, but the right triangles not isosceles.

enter image description here

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Here is what happens when the exterior point is moved to $(12,4),$ and the green circle becomes $(x-6)^2 + (y-2)^2 = 40.$ Notice that $12^2 + 4^2 > 2 \cdot 65.$

enter image description here

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    $\begingroup$ +1 for the nice geometric construction, although not a complete answer. $\endgroup$ – Dylan Dec 13 '17 at 2:28
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Hint: Through implicit differentiation, we find that the slope of the tangent line at any point $(x,y)$ on the circle is $-x/y$. Therefore the tangent point must satisfy

$$ -\frac{x}{y} = \frac{y-3}{x-11} $$ $$ x^2 + y^2 = 65 $$

You can solve this sytem of equations to find the two tangent points.

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Let the point that the tangent touches the circle be $(p,q)$. The gradient of the tangent would be $\frac{q-3}{p-11}.$ and the equation is

Since tangent is perpendicular to the radius, we have

$$\frac{q}{p}\cdot \frac{q-3}{p-11}=-1$$

Along with $$p^2+q^2=65$$

solve for $p$ and $q$.

The tangent line would satisfies: $$\frac{y-3}{x-11}=\frac{q-3}{p-11}$$

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Using homogeneous coordinates, tangent lines $\mathbf l =[\lambda,\mu,\tau]^T$ to the circle satisfy $\mathbf l^TC^{-1}\mathbf l=0$, where $C$ is the dual conic $\operatorname{diag}(65,65,1)$, i.e., the coefficients of the tangent line equation $\lambda x+\mu y+\tau=0$ satisfy $65\lambda^2+65\mu^2=\tau^2$. The lines must pass through the point $(11,3)$, so we must also have $11\lambda+3\mu+\tau=0$. Since the lines don’t pass through the origin, the constant term in their equations is nonzero, so we can set $\tau=1$ and then solve the resulting system of equations for $\lambda$ and $\mu$.

In a related method, the polar line to the point $(11,3)$ intersects the circle at the point of tangency. This line is $\operatorname{diag}(1,1,-65)\cdot(11,3,1)^T$, i.e., $11x+3y=65$. Compute the intersection of this line with the circle and then derive the equations of the lines through the two intersection points and $(11,3)$. Note that both methods essentially involve finding the intersection of a conic and a line.

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[$r (= \sqrt {65})$, the radius] = [$|\frac {c}{1 + m^2}|$, the distance of the center (0, 0) from the tangent (y = mx + c)].

Then, $|\frac {c}{\sqrt {1 + m^2}}| = r$. That is, $c^2 = r^2(1 + m^2)$.

Therefore, the equation of tangent is $y = mx \pm r\sqrt(1 + m^2)$.

Plug (11, 3) into the last equation to get the values of m. The corresponding values of c can be found accordingly.

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