Find the equations of tangents drawn from point $(11,3)$ to the circle $x^2+y^2=65$
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.
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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 $12^2 + 4^2 > 2 \cdot 65.$
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Updated on August 01, 2022Comments
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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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Dylan almost 6 years+1 for the nice geometric construction, although not a complete answer.