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

1,342

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

....................................

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

Share:
1,342

Related videos on Youtube

pi-π
Author by

pi-π

Updated on August 01, 2022

Comments

  • pi-π
    pi-π over 1 year

    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 $4x+7y-65=0$.

  • Dylan
    Dylan almost 6 years
    +1 for the nice geometric construction, although not a complete answer.