Question

. Let two circles C1 and C2 intersect at X and Y. Prove that a point P is on the line XY if and only if the power of P with respect to C1 is equal to the power of P with respect to C

Answer 1

The power of a point theorem, due to Jakob Steiner, states that for any line through A intersecting a circle c in points P and Q, the power of the point with respect to the circle c is given up to a sign by the product

AP.AQ

of the lengths of the segments from A to P and A to Q, with a positive sign if A is outside the circle and a negative sign otherwise: if A is on the circle, the product is zero. In the limiting case, when the line is tangent to the circle, P = Q, and the result is immediate from the Pythagorean theorem.

Answer to the question

Power of p with respect to c1 =(PO₁)²-r²

=PX.PY ( By the definition of

Power of the point)

Power of p with respect to c2= (PO₂)²-r²

=PX.PY

PO = distance between point p and center of circle.

r = radius of circle

Therefore power of P w.r.t to c1= power of P w.r.t to c2

That implies  a point P is on the line XY if and only if the power of P with respect to C1 is equal to the power of P with respect to C.

For your understanding of power of a point
Suppose on a plane we have a point P and a circle (O). Draw a line through P which intersects with the circle at two points U and V. Then the value of

PU×PV

is independent of the choice of the line PUV.

This means that if we draw another line through P which cuts the circle at two other points A and B then

PA×PB=PU×PV.

This constant value is called the power of the point P with respect to the circle (O).

To prove that PU×PV is independent of the position of the line, we will use similar triangles. We will consider two separate cases: when the point P is outside the circle and when the point P is inside the circle.


When P is outside the circle (O)

We consider two triangles PUB and PAV. These two triangles are similar triangles because they have two pairs of equal angles. Thus,

PU/PA=PB/PV

From here we obtain the required equality

PA×PB=PU×PV.

When P is inside the circle (O)

When P is inside the circle, we also consider the two triangles PUB and PAV. These two triangles are similar triangles because they have equal angles. So

PU/PA=PB/PV.

This implies the constancy of the power of the point P:

PA×PB=PU×PV

Also, another definition way to calculate

Power of the point = PO²​​​​ - r²

^{where PO is the distance between the point p and the centre of circle o, and r is the radius of the circle}