two circle geometry problem?

Question

Hi, I need a little help with this problem...

Two circles α and β meet at P and Q. PS and PT are diameters of the circles. Show that S, Q and T are collinear.

Answer 1

It isn't too hard, if you remember that an angle with vertex on a circle has measure 1/2 the intercepted arc. (or, a specific case, a circumscribed triangle with 1 side a diameter is a right triangle)

Draw QS and QT. Then angle SQP is 90 degrees, since the intercepted arc is 180 (because SP is a diameter.

Similarly, PQT is 90 degrees. So angle SQT = SQP + PQT = 90 + 90 = 180, so SQT is a line segment.

Answer 2

You have two inscribed triangles, one in each circle. They are both right triangles because, one side is the diameter. That means the opposing side from the diameter is the right angle. So the two triangles both have a right angle at Point Q. The angles are

Answer 3

You and mathemathe must be in the same class. As I said in my answer to her, these are hard to do without diagrams. So, I'll just outline the proof for you:

Define the centers of the circle, say O1 and O2.

Use the fact that the sum of the angles of a triangle is 180.

Use the fact that isosceles triangles have two congruent angles.

Match up the congruent angles to show SQT is a "straight angle", i.e. the angle is 180 degrees. That proves S, Q and T are collinear.


I can clarify this a little by saying I see four angles around point Q that are congruent with angles in triangle PST.

Answer 4

r u in class x. if yes, then hi, im also in class x

the proof is easy, in segment PQT, PT is dia., so angle PQT is a right angle. similarily, angle PQS is also right angle.
they form a linear pair.
=> SQT is a straight line and all 3 points lie on the same line.

Answer 5

permit Al and As be the element of two circles and additionally permit Rl and Rs be the radius respectively.. Al/As = 9/4 = (pi)(Rl)^2 / (pi)(Rs)^2 Rl/Rs = 3/2 or Rl = 3/2Rs = a million a million/2Rs <= ansb)