I draw the tangent from point A along circle B (I calculate it through the intersection with circle C), I get the coordinates D, E. How can I find points F and G? The radius of the circle marked with a dotted line (the points F and G lie on it) is also known. The large circle marked with a bold dotted line also has a center at point A

What formula is used?

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  • The formula used is that the triangle inscribed in a circle and based on its diameter is rectangular. - Igor
  • @igor can tell how to use it? - Depish

2 answers 2

The triangle ABD is rectangular. You know the radius of the circle B and the length AD , that is, you can find any angle in ABD and the coordinates of point D If we assume that the origin of coordinates at point А , then the coordinates of point F , are proportional to the coordinates of point D , with the coefficient |AF|/|AD| .

  • it says exactly - AR Hovsepyan
  • ABD do not form a right angle! point D is a tangent from point A to circle B and this point will always change depending on the distance - Depish
  • 2
    the point changes and the angle AD relative to AB changes too, and the angle ADB will always be straight - AR Hovsepyan
  • one
    @eSkry A tangent to a circle is always perpendicular to the radius drawn to the point of tangency. Euclide was not a fraer. - Igor

I suggest going through vectors. Below we use the rectangular Cartesian coordinate system with the origin at the point A(0; 0) , the horizontal axis Ax and the vertical axis Ay .

Vector АF = k vector AD . Vector АG = k vector AE .

Vector AD = (xD - xA; yD - yA) = (xD, yD ). Vector AE = (xE - xA, yE - yA) = (xE, yE) .

The coefficient of proportionality k is equal to the ratio of the lengths of the vectors AF and AD , where the length AF is equal to the known radius R circle shown in bold dotted lines. And the length of the vector AD is equal to the square root of the sum of the squares of the coordinates of this vector, that is, the length AD = (xD^2 + yD^2)^0.5 .

Knowing k , we find AF = (k*xD; k*yD) .

Knowing the coordinates of the vector AF and the coordinates of its beginning And you can find the coordinates of its end, adding the corresponding coordinates of the vector and the beginning. Given that the beginning of the vector coincides with the origin, we conclude that the point F has coordinates (k*xD; k*yD) , where k=R/(xD^2 + yD^2)^0.5

Similarly, G has coordinates (k*xG, k*yG) .

  • I don't see something, where are the explanations of how to find xD and yD . - Igor
  • @Igor The questioner in the question indicates "I get the coordinates D, E". Accordingly, it is not asked. I presume that this is known. - iramm
  • Is logical. I wonder how he finds the coordinates of point D , not counting the triangle ABD rectangular. - Igor
  • @Igor From his words it can be assumed that he solves a system of equations consisting of equations of two circles (with center at points B and D, respectively). - iramm